陶哲轩Analysis I习题的参考解答及思考（第7章）

第7章

版本

Analysis I（第3版）。

定理：基础情况从任意整数开始的数学归纳法

正常情况下，遇到基础情况不从0开始的情况，我使用数学归纳会对“位移”变量进行数学归纳，比如基础情况从\( 1 \)开始，我会设\( m = n - 1 \)，从而将基础情况变成从\( 0 \)开始，变得与数学归纳法公理框架一致（可以去看前几章的参考解答），但是实际上基数情况从任意整数开始都可以，设下对应的位移变量，比如基础情况从\( n = k \)开始（不管\( k > 0 \)也好，\( k < 0 \)也好），设\( m = n - k \)，都能变成基础从0开始的数学归纳法，最终得出的结论也会变成\( n \geq k \)（当然，这要求用到序，然而序早就定义完了）。

但是每次都设位移变量太麻烦了（还有变量回代），直接基础情况从\( n = k \) 开始证是最简单的，为此我们证明如下的命题：

Let \( k \) be an integer, let \( P(n) \) be any property pertaining to an integer \( n \geq k \). Suppose that \( P(k) \) is true, and suppose that whenever \( P(n) \) is true for an integer \( n \geq k \), \( P(n+\!+) \) is also true. Then \( P(n) \) is true for every integers \( n \geq k \).

附注1：

\( P(n) \Rightarrow P(n+\!+) \)中，\( n \)为\( \geq k \)的整数，而不是任意自然数。

附注2：

该命题和新增的练习2.2.7很像，但是我们的\( k \)可以是任意整数，而没有限定为自然数，但证明是几乎一样的。

证明：

令属性\( Q(m) \)为\( P(m + k) \)。

将\( Q(m) \)代入数学归纳法公理框架，形式如下：

首先验证/证明\( Q(0) \)为真。接着，归纳假设\( Q(m) \)为真，在此基础上，若能证明\( Q(m+\!+) \)为真，那么\( Q(m) \)对所有\( m \in \mathbf{N} \)都成立。

首先验证/证明基础情况\( Q(0) \)为真，即验证/证明\( P(0 + k) = P(k) \)为真。接着，归纳假设\( Q(m) \)为真，这里\( Q(m) \)为真即\( P(m + k) \)为真。在此基础上，若能证明\( Q(m+\!+) = P((m+\!+) + k) = P((m + k)+\!+) \)为真，则\( Q(m) = P(m + k) \)对所有\( m \in \mathbf{N} \)都成立，换句话来说就是\( \forall \)整数\( k' \geq k \)，均有\( P(k') \)为真。

综上，如果能首先验证/证明\( P(k) \)为真，且在假设给定任意整数\( m \geq k \)，\( P(m) \)为真的基础上，能推出\( P(m+\!+) \)为真，则\( P(m) \)对所有整数\( m \geq k \)都成立。

证毕。

故从这里开始，我们可以以任意整数作为基础情况进行数学归纳。

附注

第3章中，我证明了很多关于函数是否单射、满射、双射的练习，大部分证明都没有什么难度，也比较无聊，故本章中，当遇到要证明某个函数为双射函数时，如果我 觉得 比较容易证明时，我可能会写易证。

章节7.1

练习7.1.1

题目：

Prove Lemma 7.1.4. (Hint: you will need to use induction, but the base case might not necessarily be at 0.)

Lemma 7.1.4的内容：

1. Let \( m \leq n < p \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \leq i \leq p \). Then we have \( \sum\limits_{i = m}^{n} a_i + \sum\limits_{i = n + 1}^{p} a_i = \sum\limits_{i = m}^{p} a_i \).
2. Let \( m \leq n \) be integers, \( k \) be another integer, and let \( a_i \) be a real number assigned to each integer \( m \leq i \leq n \). Then we have \( \sum\limits_{i = m}^{n} a_i = \sum\limits_{j = m + k}^{n + k} a_{j - k} \).
3. Let \( m \leq n \) be integers, and let \( a_i, b_i \) be real numbers assigned to each integer \( m \leq i \leq n \). Then we have \( \sum\limits_{i = m}^{n} (a_i + b_i) = (\sum\limits_{i = m}^{n} a_i) + (\sum\limits_{i = m}^{n} b_i) \).
4. Let \( m \leq n \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \leq i \leq n \), and let \( c \) be another real number. Then we have \( \sum\limits_{i = m}^{n} (ca_i) = c(\sum\limits_{i = m}^{n} a_i) \).
5. (Triangle inequality for finite series) Let \( m \leq n \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \leq i \leq n \). Then we have \( |\sum\limits_{i = m}^{n} a_i| \leq \sum\limits_{i = m}^{n} |a_i| \).
6. (Comparison test for finite series) Let \( m \leq n \) be integers, and let \( a_i, b_i \) be real numbers assigned to each integer \( m \leq i \leq n \). Suppose that \( a_i \leq b_i \) for all \( m \leq i \leq n \). Then we have \( \sum\limits_{i = m}^{n} a_i \leq \sum\limits_{i = m}^{n} b_i \).

• 命题1

  证明：

  对\( p \)进行数学归纳，因为\( p > n \)，有\( p \geq n + 1 \)，故基础情况从\( n + 1 \)开始。

  当\( p = n + 1 \)时， \( \sum\limits_{i = m}^{n} a_i + \sum\limits_{i = n + 1}^{p} a_i = \sum\limits_{i = m}^{n} a_i + \sum\limits_{i = n + 1}^{n + 1} a_i = (\sum\limits_{i = m}^{n} a_i) + a_{n + 1} = \sum\limits_{i = m}^{n + 1} a_i = \sum\limits_{i = m}^{p} a_i \)，即\( p = n + 1 \)时，成立。

  归纳假设当\( p = k \geq n + 1 \)时成立，即\( \sum\limits_{i = m}^{n} a_i + \sum\limits_{i = n + 1}^{k} a_i = \sum\limits_{i = m}^{k} a_i \)，当\( p = k+\!+ \)时，根据归纳假设，有 \( \sum\limits_{i = m}^{n} a_i + \sum\limits_{i = n + 1}^{p} a_i = \sum\limits_{i = m}^{n} a_i + \sum\limits_{i = n + 1}^{k+\!+} a_i = (\sum\limits_{i = m}^{n} a_i + \sum\limits_{i = n + 1}^{k} a_i) + a_{k+\!+} = (\sum\limits_{i = m}^{k} a_i) + a_{k+\!+} = \sum\limits_{i = m}^{k+\!+} a_i = \sum\limits_{i = m}^{p} a_i \)，即\( p = k+\!+ \)时成立。

  至此，归纳完毕，命题成立。

  证毕。

• 命题2

  证明：

  对\( n \)进行数学归纳，基础情况从\( m \)开始。

  当\( n = m \)时，左边 = \( \sum\limits_{i = m}^{n} a_i = \sum\limits_{i = m}^{m} a_i = a_m \)，右边 = \( \sum\limits_{j = m + k}^{n + k} a_{j - k} = \sum\limits_{j = m + k}^{m + k} a_{j - k} = a_{m + k - k} = a_m \)，左边 = 右边，即基础情况下成立。

  归纳假设当\( n = c \geq m \)时成立，即\( \sum\limits_{i = m}^{c} a_i = \sum\limits_{j = m + k}^{c + k} a_{j - k} \)，当\( n = c+\!+ \)时，根据归纳假设，有 \( \sum\limits_{i = m}^{n} a_i = \sum\limits_{i = m}^{c+\!+} a_i = (\sum\limits_{i = m}^{c} a_i) + a_{c+\!+} = (\sum\limits_{j = m + k}^{c + k} a_{j - k}) + a_{c+\!+} = (\sum\limits_{j = m + k}^{c + k} a_{j - k}) + a_{(c + k + 1) - k} = \sum\limits_{j = m + k}^{c + k + 1} a_{j - k} \)，即\( n = c+\!+ \)时成立。

  至此，归纳完毕，命题成立。

  证毕。

• 命题3

  证明：

  对\( n \)进行数学归纳，基础情况从\( m \)开始。

  当\( n = m \)时，左边 = \( \sum\limits_{i = m}^{n} (a_i + b_i) = \sum\limits_{i = m}^{m} (a_i + b_i) = a_m + b_m \)，右边 = \( (\sum\limits_{i = m}^{n} a_i) + (\sum\limits_{i = m}^{n} b_i) = (\sum\limits_{i = m}^{m} a_i) + (\sum\limits_{i = m}^{m} b_i) = a_m + b_m \)，左边 = 右边，即基础情况下成立。

  归纳假设当\( n = k \geq m \)时成立，即\( \sum\limits_{i = m}^{k} (a_i + b_i) = (\sum\limits_{i = m}^{k} a_i) + (\sum\limits_{i = m}^{k} b_i) \)，当\( n = k+\!+ \)时，根据归纳假设，有 \( \sum\limits_{i = m}^{n} (a_i + b_i) = \sum\limits_{i = m}^{k+\!+} (a_i + b_i) = (\sum\limits_{i = m}^{k} (a_i + b_i)) + (a_{k+\!+} + b_{k+\!+}) = ((\sum\limits_{i = m}^{k} a_i) + (\sum\limits_{i = m}^{k} b_i)) + (a_{k+\!+} + b_{k+\!+}) = ((\sum\limits_{i = m}^{k} a_i) + a_{k+\!+}) + ((\sum\limits_{i = m}^{k} b_i) + b_{k+\!+}) = (\sum\limits_{i = m}^{k+\!+} a_i) + (\sum\limits_{i = m}^{k+\!+} b_i) = (\sum\limits_{i = m}^{n} a_i) + (\sum\limits_{i = m}^{n} b_i) \)，即\( n = k+\!+ \)时成立。

  至此，归纳完毕，命题成立。

  证毕。

• 命题4

  证明：

  对\( n \)进行数学归纳，基础情况从\( m \)开始。

  当\( n = m \)时，左边 = \( \sum\limits_{i = m}^{n} (ca_i) = \sum\limits_{i = m}^{m} (ca_i) = ca_m \)，右边 = \( c(\sum\limits_{i = m}^{n} a_i) = c(\sum\limits_{i = m}^{m} a_i) = ca_m \)，左边 = 右边，即基础情况下成立。

  归纳假设当\( n = k \geq m \)时成立，即\( \sum\limits_{i = m}^{k} (ca_i) = c(\sum\limits_{i = m}^{k} a_i) \)，当\( n = k+\!+ \)时，根据归纳假设，有 \( \sum\limits_{i = m}^{n} (ca_i) = \sum\limits_{i = m}^{k+\!+} (ca_i) = (\sum\limits_{i = m}^{k} (ca_i)) + ca_{k+\!+} = (c(\sum\limits_{i = m}^{k} a_i)) + ca_{k+\!+} = c((\sum\limits_{i = m}^{k} a_i) + a_{k+\!+}) = c(\sum\limits_{i = m}^{k+\!+} a_i) = c(\sum\limits_{i = m}^{n} a_i) \)，即\( n = k+\!+ \)时成立。

  至此，归纳完毕，命题成立。

  证毕。

• 命题5

  证明：

  对\( n \)进行数学归纳，基础情况从\( m \)开始。

  当\( n = m \)时，左边 = \( |\sum\limits_{i = m}^{n} a_i| = |\sum\limits_{i = m}^{m} a_i| = |a_m| \)，右边 = \( \sum\limits_{i = m}^{n} |a_i| = \sum\limits_{i = m}^{m} |a_i| = |a_m| \)，左边\( \leq \)右边，即基础情况下成立。

  归纳假设当\( n = k \geq m \)时成立，即\( |\sum\limits_{i = m}^{k} a_i| \leq \sum\limits_{i = m}^{k} |a_i| \)，当\( n = k+\!+ \)时，根据归纳假设以及绝对值不等式，有\( |\sum\limits_{i = m}^{n} a_i| = |\sum\limits_{i = m}^{k+\!+} a_i| = |(\sum\limits_{i = m}^{k} a_i) + a_{k+\!+}| \leq |(\sum\limits_{i = m}^{k} a_i)| + |a_{k+\!+}| \leq \sum\limits_{i = m}^{k} |a_i| + |a_{k+\!+}| = \sum\limits_{i = m}^{k+\!+} |a_i| = \sum\limits_{i = m}^{n} |a_i| \)，即\( n = k+\!+ \)时成立。

  至此，归纳完毕，命题成立。

  证毕。

• 命题6

  证明：

  对\( n \)进行数学归纳，基础情况从\( m \)开始。

  当\( n = m \)时，如果\( \forall m \leq i \leq n \)，有\( a_i \leq b_i \)，此时，左边 = \( \sum\limits_{i = m}^{n} a_i = \sum\limits_{i = m}^{m} a_i = a_m \)，右边 = \( \sum\limits_{i = m}^{n} b_i = \sum\limits_{i = m}^{m} b_i = b_m \)，又\( a_m \leq b_m \)，可得左边\( \leq \)右边，即基础情况下成立。

  归纳假设当\( n = k \geq m \)时成立，即如果\( \forall m \leq i \leq k \)，有\( a_i \leq b_i \)，则\( \sum\limits_{i = m}^{k} a_i \leq \sum\limits_{i = m}^{k} b_i \)，当\( n = k+\!+ \)时，如果\( \forall m \leq i \leq k+\!+ \)，有\( a_i \leq b_i \)，根据归纳假设，有\( \sum\limits_{i = m}^{k} a_i \leq \sum\limits_{i = m}^{k} b_i \)，又\( a_{k+\!+} \leq b_{k+\!+} \)，可得\( (\sum\limits_{i = m}^{k} a_i) + a_{k+\!+} \leq (\sum\limits_{i = m}^{k} b_i) + b_{k+\!+} \)，即\( \sum\limits_{i = m}^{k+\!+} a_i \leq \sum\limits_{i = m}^{k+\!+} b_i \)，有\( n = k+\!+ \)时成立。

  至此，归纳完毕，命题成立。

  证毕。

练习7.1.2

题目：

Prove Proposition 7.1.11. (Hint: this is not as lengthy as it may first appear. It is largely a matter of choosing the right bijections to turn these sums over sets into finite series, and then applying Lemma 7.1.4.)

Proposition 7.1.11的内容：

1. If \( X \) is empty, and \( f: X \to \mathbf{R} \) is a function (i.e., \( f \) is the empty function), we have \( \sum\limits_{x \in X} f(x) = 0 \).
2. If \( X \) consists of a single element, \( X = \{ x_0 \} \), and \( f: X \to \mathbf{R} \) is a function, we have \( \sum\limits_{x \in X} f(x) = f(x_0) \).
3. (Substitution, part I) If \( X \) is a finite set, \( f: X \to \mathbf{R} \) is a function, and \( g: Y \to X \) is a bijection, then \( \sum\limits_{x \in X} f(x) = \sum\limits_{y \in Y} f(g(y)) \).
4. (Substitution, part II) Let \( n \leq m \) be integers, and let \( X \) be the set \( X := \{ i \in \mathbf{Z} : n \leq i \leq m \} \). If \( a_i \) is a real number assigned to each integer \( i \in X \), then we have \( \sum\limits_{i = n}^{m} a_i = \sum\limits_{i \in X} a_i \).
5. Let \( X, Y \) be disjoint finite sets (so \( X \cap Y = \emptyset \)), and \( f: X \cup Y \to \mathbf{R} \) is a function. Then we have \( \sum\limits_{z \in X \cup Y} f(z) = (\sum\limits_{x \in X} f(x)) + (\sum\limits_{y \in Y} f(y)) \).
6. (Linearity, part I) Let \( X \) be a finite set, and let \( f: X \to \mathbf{R} \) and \( g: X \to \mathbf{R} \) be functions. Then \( \sum\limits_{x \in X} (f(x) + g(x)) = \sum\limits_{x \in X} f(x) + \sum\limits_{x \in X} g(x) \).
7. (Linearity, part II) Let \( X \) be a finite set, let \( f: X \to \mathbf{R} \) be a function, and let \( c \) be a real number. Then \( \sum\limits_{x \in X} cf(x) = c(\sum\limits_{x \in X} f(x)) \).
8. (Monotonicity) Let \( X \) be a finite set, and let \( f: X \to \mathbf{R} \) and \( g: X \to \mathbf{R} \) be functions such that \( f(x) \leq g(x) \) for all \( x \in X \). Then we have \( \sum\limits_{x \in X} f(x) \leq \sum\limits_{x \in X} g(x) \).
9. (Triangle inequality) Let \( X \) be a finite set, and let \( f: X \to R \) be a function, then \( |\sum\limits_{x \in X} f(x)| \leq \sum\limits_{x \in X} |f(x)| \).

• 命题1

  证明：

  \( \forall g: \{ i \in \mathbf{N} : 1 \leq i \leq 0 \} \to X \)，因为\( X = \emptyset \)，有\( g \)为双射函数，此时\( \sum\limits_{x \in X} f(x) = \sum\limits_{i = 1}^{0} f(g(i)) = 0 \)。

  证毕。

• 命题2

  证明：

  构造函数\( g: \{ i \in \mathbf{N} : 1 \leq i \leq 1 \} \to X \)，令\( g(1) = x_0 \)，易证\( g \)为双射函数，此时\( \sum\limits_{x \in X} f(x) = \sum\limits_{i = 1}^{1} f(g(i)) = f(g(1)) = f(x_0) \)。

  证毕。

• 命题3

  证明：

  因为\( X \)有限，记它的基数为\( n \)，可得\( \exists \)双射函数\( h: \{ i \in \mathbf{N} : 1 \leq i \leq n \} \to X \)，又\( g: Y \to X \)为双射函数，有\( g^{-1} \circ h \)为\( \{ i \in \mathbf{N} : 1 \leq i \leq n \} \to Y \)的双射函数，此时左边 = \( \sum\limits_{x \in X} f(x) = \sum\limits_{i = 1}^{n} f(h(i)) \)，右边 = \( \sum\limits_{y \in Y} f(g(y)) = \sum\limits_{i = 1}^{n} f(g((g^{-1} \circ h)(i))) = \sum\limits_{i = 1}^{n} f(g(g^{-1}(h(i)))) = \sum\limits_{i = 1}^{n} f(h(i)) \)，可得左边 = 右边。

  证毕。

• 命题4

  证明：

  对\( m \)进行数学归纳，基础情况从\( n \)开始。

  当\( m = n \)时，\( X = \{ n \} \)只有一个元素，左边 = \( \sum\limits_{i = n}^{m} = \sum\limits_{i = n}^{n} = a_n \)，根据命题2，右边 = \( a_n \)（注意：\( a \)实际上是个函数，\( a_n \)只是\( a(n) \)的另一种记法），左边 = 右边，即基础情况下成立。

  归纳假设当\( m = k \geq n \)时成立，即\( X = \{ i \in \mathbf{Z} : n \leq i \leq k \} \)， \( \sum\limits_{i = n}^{k} a_i = \sum\limits_{i \in X} a_i \)。当\( m = k+\!+ \)时，\( X = \{ i \in \mathbf{Z} : n \leq i \leq k+\!+ \} \)，令\( Y = X \setminus \{ k+\!+ \} = \{ i \in \mathbf{Z} : n \leq i \leq k \} \)， \( Y \)基数为\( k - n + 1 \)，故\( \exists \)双射函数 \( g: \{ i \in \mathbf{N} : 1 \leq i \leq k - n + 1 \} \to Y \)，此时\( \sum\limits_{i \in Y} a_i = \sum\limits_{i = 1}^{k - n + 1} a_{g(i)} \)（或者写成\( a(g(i)) \)也行），根据归纳假设，有\( \sum\limits_{i = n}^{k} a_i = \sum\limits_{i = 1}^{k - n + 1} a_{g(i)} \)，构造函数\( g_1: \{ i \in \mathbf{N} : 1 \leq i \leq k - n + 2 \} \to X \)， \( \forall 1 \leq i \leq k - n + 2 \)，如果\( i < k - n + 2 \)，则\( g_1(i) = g(i) \)，如果\( i = k - n + 2 \)，则\( g_1(i) = k+\!+ \)，易证\( g_1 \)为双射函数，此时左边 = \( \sum\limits_{i = n}^{k+\!+} a_i \)，右边 = \( \sum\limits_{i \in X} a_i = \sum\limits_{i = 1}^{k - n + 2} a_{g_1(i)} = (\sum\limits_{i = 1}^{k - n + 1} a_{g_1(i)}) + a_{g_1(k - n + 2)} = (\sum\limits_{i = 1}^{k - n + 1} a_{g_1(i)}) + a_{k+\!+} = (\sum\limits_{i = 1}^{k - n + 1} a_{g(i)}) + a_{k+\!+} = (\sum\limits_{i = n}^{k} a_i) + a_{k+\!+} = \sum\limits_{i = n}^{k+\!+} a_i \)，左边 = 右边，即\( m = k+\!+ \)时成立。

  至此，归纳完毕，命题成立。

  证毕。

• 命题5

  证明：

  记\( X \)的基数为\( m \)，\( Y \)的基数为\( n \)，对\( X \)的基数\( m \)进行数学归纳。

  当\( m = 0 \)时，\( X = \emptyset \)，\( X \cup Y = Y \)，根据命题1，有\( \sum\limits_{x \in X} f(x) = 0 \)，此时，左边 = \( \sum\limits_{z \in X \cup Y} f(z) = \sum\limits_{y \in Y} f(y) \)，右边 = \( (\sum\limits_{x \in X} f(x)) + (\sum\limits_{y \in Y} f(y)) = (\sum\limits_{x \in X} f(x)) + (\sum\limits_{y \in Y} f(y)) = 0 + (\sum\limits_{y \in Y} f(y)) = \sum\limits_{y \in Y} f(y) \)，左边 = 右边，即基础情况下成立。

  归纳假设当\( m = k \)时成立，当\( m = k+\!+ \)时，因为\( X \)非空，故\( \exists x_0 \in X \)，此时令\( Z = X \setminus \{ x_0 \} \)，可得\( Z \)的基数为\( k \)，因为\( X \cap Y = \emptyset \)，可得\( Z \cap Y = \emptyset \)，根据定理3.6.14的命题2，有\( Z \cup Y \)的基数为\( k + n \)，\( X \cup Y \)的基数为\( k + n + 1 \)；可得存在双射函数\( g_0: \{ i \in \mathbf{N} : 1 \leq i \leq k + n \} \to Z \cup Y \)，使得\( \sum\limits_{z \in Z \cup Y} f(z) = \sum\limits_{i = 1}^{k + n} f(g_0(i)) \)；存在双射函数\( g_1: \{ i \in \mathbf{N} : 1 \leq i \leq k \} \to Z \)，使得\( \sum\limits_{x \in Z} f(x) = \sum\limits_{i = 1}^{k} f(g_1(i)) \)；存在双射函数\( g_2: \{ i \in \mathbf{N} : 1 \leq i \leq n \} \to Y \)，使得\( \sum\limits_{y \in Y} f(y) = \sum\limits_{i = 1}^{n} f(g_2(i)) \)；根据归纳假设，有\( \sum\limits_{z \in Z \cup Y} f(z) = (\sum\limits_{x \in Z} f(x)) + (\sum\limits_{y \in Y} f(y)) \)，即\( \sum\limits_{i = 1}^{k + n} f(g_0(i)) = (\sum\limits_{i = 1}^{k} f(g_1(i))) + (\sum\limits_{i = 1}^{n} f(g_2(i))) \)；构造函数\( g_3: \{ i \in \mathbf{N} : 1 \leq i \leq k + n + 1 \} \to X \cup Y \)， \( \forall 1 \leq i \leq k + n + 1 \)，如果\( i < k + n + 1 \)，则\( g_3(i) = g_0(i) \)，如果\( i = k + n + 1 \)，则\( g_3(i) = x_0 \)，易证\( g_3 \)为双射函数；构造函数\( g_4: \{ i \in \mathbf{N} : 1 \leq i \leq k + 1 \} \to X \)， \( \forall 1 \leq i \leq k + 1 \)，如果\( i < k + 1 \)，则\( g_4(i) = g_1(i) \)，如果\( i = k + 1 \)，则\( g_4(i) = x_0 \)，易证\( g_4 \)为双射函数；此时，左边 = \( \sum\limits_{z \in X \cup Y} f(z) = \sum\limits_{i = 1}^{k + n + 1} f(g_3(i)) = (\sum\limits_{i = 1}^{k + n} f(g_3(i))) + f(g_3(k + n + 1)) = (\sum\limits_{i = 1}^{k + n} f(g_3(i))) + f(x_0) = (\sum\limits_{i = 1}^{k + n} f(g_0(i))) + f(x_0) = ((\sum\limits_{i = 1}^{k} f(g_1(i))) + (\sum\limits_{i = 1}^{n} f(g_2(i)))) + f(x_0) \)，右边 = \( (\sum\limits_{i = 1}^{k + 1} f(g_4(i))) + (\sum\limits_{i = 1}^{n} f(g_2(i))) = ((\sum\limits_{i = 1}^{k} f(g_4(i))) + f(g_4(k + 1))) + (\sum\limits_{i = 1}^{n} f(g_2(i))) = ((\sum\limits_{i = 1}^{k} f(g_4(i))) + f(x_0)) + (\sum\limits_{i = 1}^{n} f(g_2(i))) = ((\sum\limits_{i = 1}^{k} f(g_4(i))) + (\sum\limits_{i = 1}^{n} f(g_2(i)))) + f(x_0) \)，左边 = 右边，即\( m = k+\!+ \)时成立。

  至此，归纳完毕，命题成立。

  证毕。

• 命题6

  证明：

  记\( X \)的基数为\( n \)，对\( n \)进行数学归纳。

  当\( n = 0 \)时，\( X = \emptyset \)，根据命题1，左边 = \( \sum\limits_{x \in X} (f(x) + g(x)) = 0 \)，右边 = \( (\sum\limits_{x \in X} f(x)) + (\sum\limits_{x \in X} g(x)) = 0 + 0 = 0 \)，左边 = 右边，即基础情况下成立。

  归纳假设成\( n = k \)时成立，当\( n = k+\!+ \)时，因为\( X \)非空，故\( \exists x_0 \in X \)，令\( Y = X \setminus \{ x_0 \} \)，可得\( Y \)的基数为\( k \)，故存在双射函数\( h: \{ i \in \mathbf{N} : 1 \leq i \leq k \} \to Y \)，使得\( \sum\limits_{x \in X} (f(x) + g(x)) = \sum\limits_{i = 1}^{k} (f(h(i)) + g(h(i))) \)， \( \sum\limits_{x \in X} f(x) = \sum\limits_{i = 1}^{k} f(h(i)) \)， \( \sum\limits_{x \in X} g(x) = \sum\limits_{i = 1}^{k} g(h(i)) \)，根据归纳假设，有\( \sum\limits_{x \in X} (f(x) + g(x)) = \sum\limits_{x \in X} f(x) + \sum\limits_{x \in X} g(x) \)，即\( \sum\limits_{i = 1}^{k} (f(h(i)) + g(h(i))) = (\sum\limits_{i = 1}^{k} f(h(i))) + (\sum\limits_{i = 1}^{k} g(h(i))) \)，构造函数\( h_1: \{ i \in \mathbf{N} : 1 \leq i \leq k + 1 \} \to X \)， \( \forall 1 \leq i \leq k + 1 \)，如果\( i < k + 1 \)，则\( h_1(i) = h(i) \)，如果\( i = k + 1 \)，则\( h_1(i) = x_0 \)，易证\( h_1 \)为双射函数，此时左边 = \( \sum\limits_{x \in X} (f(x) + g(x)) = \sum\limits_{i = 1}^{k + 1} (f(h_1(i)) + g(h_1(i))) = (\sum\limits_{i = 1}^{k} (f(h_1(i)) + g(h_1(i)))) + (f(h_1(k + 1)) + g(h_1(k + 1))) = (\sum\limits_{i = 1}^{k} (f(h_1(i)) + g(h_1(i)))) + (f(x_0) + g(x_0)) = ((\sum\limits_{i = 1}^{k} f(h(i))) + (\sum\limits_{i = 1}^{k} g(h(i)))) + (f(x_0) + g(x_0)) \)，右边 = \( \sum\limits_{x \in X} f(x) + \sum\limits_{x \in X} g(x) = (\sum\limits_{i = 1}^{k + 1} f(h_1(i))) + (\sum\limits_{i = 1}^{k + 1} g(h_1(i))) = ((\sum\limits_{i = 1}^{k} f(h_1(i))) + f(h_1(k + 1))) + ((\sum\limits_{i = 1}^{k + 1} g(h_1(i))) + g(h_1(k + 1))) = ((\sum\limits_{i = 1}^{k} f(h_1(i))) + f(x_0)) + ((\sum\limits_{i = 1}^{k + 1} g(h_1(i))) + g(x_0)) = ((\sum\limits_{i = 1}^{k} f(h(i))) + (\sum\limits_{i = 1}^{k} g(h(i)))) + (f(x_0) + g(x_0)) \)，左边 = 右边，即\( n = k+\!+ \)时成立。

  至此，归纳完毕，命题成立。

  证毕。

• 命题7

  证明：

  记\( X \)的基数为\( n \)，可得存在双射函数\( g: \{ i \in \mathbf{N} : 1 \leq i \leq n \} \to X \)，此时，左边 = \( \sum\limits_{x \in X} cf(x) = \sum\limits_{i = 1}^{n} cf(g(i)) \)，根据引理7.1.4的命题4，有\( \sum\limits_{i = 1}^{n} cf(g(i)) = c(\sum\limits_{i = 1}^{n} f(g(i))) = c(\sum\limits_{x \in X} f(x)) \) = 右边（注：这里函数\( f \circ g \)相当于引理7.1.4的\( a \)），左边 = 右边，命题成立。

  证毕。

• 命题8

  证明：

  记\( X \)的基数为\( n \)，可得存在双射函数\( h: \{ i \in \mathbf{N} : 1 \leq i \leq n \} \to X \)，此时，左边 = \( \sum\limits_{x \in X} f(x) = \sum\limits_{i = 1}^{n} f(h(i)) \)，右边 = \( \sum\limits_{x \in X} g(x) = \sum\limits_{i = 1}^{n} g(h(i)) \)，因为\( \forall x \in X \)，有\( f(x) \leq g(x) \)，可得\( \forall 1 \leq i \leq n \)，有\( f(h(i)) \leq g(h(i)) \)，根据引理7.1.4的命题6，有\( \sum\limits_{i = 1}^{n} f(h(i)) \leq \sum\limits_{i = 1}^{n} g(h(i)) \)，即左边\( \leq \)右边，命题成立。

  证毕。

• 命题9

  证明：

  记\( X \)的基数为\( n \)，可得存在双射函数\( g: \{ i \in \mathbf{N} : 1 \leq i \leq n \} \to X \)，此时，左边 = \( |\sum\limits_{x \in X} f(x)| = |\sum\limits_{i = 1}^{n} f(g(i))| \)，根据引理7.1.4的命题5，有\( |\sum\limits_{i = 1}^{n} f(g(i))| \leq \sum\limits_{i = 1}^{n} |f(g(i))| = \sum\limits_{x \in X} |f(x)| \) = 右边，左边 = 右边，命题成立。

  证毕。

练习7.1.3

Form a definition for the finite products \( \prod_{i = 1}^{n} a_i \) and \( \prod_{x \in X} f(x) \). Which of the above results for finite series have analogues for finite products? (Note that it is dangerous to apply logarithms because some of the \( a_i \) or \( f(x) \) could be zero or negative. Besides, we haven’t defined logarithms yet.)

• 关于\( \prod_{i = 1}^{n} a_i \)

  定义：

  (Finite products). Let m, n be integers, and let \( (a_i)_{i = m}^{n} \) be a finite sequence of real numbers, assigning a real number \( a_i \) to each integer \( i \) between \( m \) and \( n \) inclusive (i.e., \( m \leq i \leq n \)). Then we define the finite products \( \prod_{i = m}^{n} a_i \) by the recursive formula:

  1. \( \prod_{i = m}^{n} a_i := 1 \) whenever \( n < m \);
  2. \( \prod_{i = m}^{n + 1} a_i := (\prod_{i = m}^{n} a_i) \times a_{n + 1} \) whenever n ≥ m - 1.

  引理7.1.4中有哪些命题在有限积对应的版本中也成立：

  注：这里我不会给出证明，故我这里说的有可能是错的。

  1. Let \( m \leq n < p \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \leq i \leq p \). Then we have \( \prod\limits_{i = m}^{n} a_i \times \prod\limits_{i = n + 1}^{p} a_i = \prod\limits_{i = m}^{p} a_i \).
  2. Let \( m \leq n \) be integers, \( k \) be another integer, and let \( a_i \) be a real number assigned to each integer \( m \leq i \leq n \). Then we have \( \prod\limits_{i = m}^{n} a_i = \prod\limits_{j = m + k}^{n + k} a_{j - k} \).
  3. Let \( m \leq n \) be integers, and let \( a_i, b_i \) be real numbers assigned to each integer \( m \leq i \leq n \). Then we have \( \prod\limits_{i = m}^{n} (a_i \times b_i) = (\prod\limits_{i = m}^{n} a_i) \times (\prod\limits_{i = m}^{n} b_i) \).
  4. Let \( m \leq n \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \leq i \leq n \), and let \( c \) be another real number. Then we have \( \prod\limits_{i = m}^{n} (ca_i) = c(\prod\limits_{i = m}^{n} a_i) \).
  5. Let \( m \leq n \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \leq i \leq n \). Then we have \( |\prod\limits_{i = m}^{n} a_i| = \prod\limits_{i = m}^{n} |a_i| \).
  6. (Comparison test for finite products) Let \( m \leq n \) be integers, and let \( a_i, b_i \) be real numbers assigned to each integer \( m \leq i \leq n \). Suppose that \( 0 \leq a_i \leq b_i \) for all \( m \leq i \leq n \). Then we have \( \prod\limits_{i = m}^{n} a_i \leq \prod\limits_{i = m}^{n} b_i \) （注意，这里我们限制\( a_i, b_i \)为非负实数）.

• 关于\( \prod_{x \in X} f(x) \)

  定义：

  (Multiplications over finite sets). Let \( X \) be a finite set with \( n \) elements (where \( n \in N \)), and let \( f: X \to \mathbf{R} \) be a function from \( X \) to the real numbers (i.e., \( f \) assigns a real number \( f(x) \) to each element \( x \) of \( X \)). Then we can define the finite sum \( \prod_{x \in X} f(x) \) as follows. We first select any bijection \( g \) from \( \{ i \in \mathbf{N} : 1 \leq i \leq n \} \) to \( X \); such a bijection exists since \( X \) is assumed to have \( n \) elements. We then define \( \prod\limits_{x \in X} f(x) := \prod\limits_{i = 1}^{n} f(g(i)) \).

  定理7.1.11中有哪些命题在有限积对应的版本中也成立：

  注：这里我不会给出证明，故我这里说的有可能是错的。

  1. If \( X \) is empty, and \( f: X \to \mathbf{R} \) is a function (i.e., \( f \) is the empty function), we have \( \prod\limits_{x \in X} f(x) = 1 \).
  2. If \( X \) consists of a single element, \( X = \{ x_0 \} \), and \( f: X \to \mathbf{R} \) is a function, we have \( \prod\limits_{x \in X} f(x) = f(x_0) \).
  3. (Substitution, part I) If \( X \) is a finite set, \( f: X \to \mathbf{R} \) is a function, and \( g: Y \to X \) is a bijection, then \( \prod\limits_{x \in X} f(x) = \prod\limits_{y \in Y} f(g(y)) \).
  4. (Substitution, part II) Let \( n \leq m \) be integers, and let \( X \) be the set \( X := \{ i \in \mathbf{Z} : n \leq i \leq m \} \). If \( a_i \) is a real number assigned to each integer \( i \in X \), then we have \( \prod\limits_{i = n}^{m} a_i = \prod\limits_{i \in X} a_i \).
  5. Let \( X, Y \) be disjoint finite sets (so \( X \cap Y = \emptyset \)), and \( f: X \cup Y \to \mathbf{R} \) is a function. Then we have \( \prod\limits_{z \in X \cup Y} f(z) = (\prod\limits_{x \in X} f(x)) \times (\prod\limits_{y \in Y} f(y)) \).
  6. (Linearity, part I) Let \( X \) be a finite set, and let \( f: X \to \mathbf{R} \) and \( g: X \to \mathbf{R} \) be functions. Then \( \prod\limits_{x \in X} (f(x) \times g(x)) = \prod\limits_{x \in X} f(x) \times \prod\limits_{x \in X} g(x) \).
  7. (Linearity, part II) Let \( X \) be a finite set, let \( f: X \to \mathbf{R} \) be a function, and let \( c \) be a real number. Then \( \prod\limits_{x \in X} cf(x) = c(\prod\limits_{x \in X} f(x)) \).
  8. (Monotonicity) Let \( X \) be a finite set, and let \( f: X \to \mathbf{R} \) and \( g: X \to \mathbf{R} \) be functions such that \( 0 \leq f(x) \leq g(x) \) for all \( x \in X \). Then we have \( \prod\limits_{x \in X} f(x) \leq \prod\limits_{x \in X} g(x) \) （注意，这里我们限制\( f(x), g(x) \)为非负实数）.
  9. Let \( X \) be a finite set, and let \( f: X \to R \) be a function, then \( |\prod\limits_{x \in X} f(x)| = \prod\limits_{x \in X} |f(x)| \).

练习7.1.4

题目：

Define the factorial function \( n! \) for natural numbers \( n \) by the recursive definition \( 0! := 1 \) and \( (n + 1)! := n! \times (n + 1) \). If \( x \) and \( y \) are real numbers, prove the binomial formula \( (x + y)^n = \sum\limits_{j = 0}^{n} \dfrac{n!}{j!(n - j)!}x^jy^{n - j} \) for all natural numbers \( n \). (Hint: induct on \( n \).)

证明：

使用数学归纳法。

当\( n = 0 \)，左边 = \( (x + y)^n = (x + y)^0 = 1 \)，右边 = \( \sum\limits_{j = 0}^{n} \dfrac{n!}{j!(n - j)!}x^jy^{n - j} = \sum\limits_{j = 0}^{0} \dfrac{0!}{j!(0 - j)!}x^0y^{0 - j} = \dfrac{0!}{0!(0 - 0)!}x^0y^{0 - 0} = 1 \)，左边 = 右边，即基础情况下成立。

归纳假设当\( n = k \)时成立，即\( (x + y)^k = \sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^jy^{k - j} \)，当\( n = k+\!+ \)时，我们要证 \( (x + y)^{k+\!+} = \sum\limits_{j = 0}^{k+\!+} \dfrac{(k+\!+)!}{j!((k+\!+) - j)!}x^jy^{(k+\!+) - j} \)，根据归纳假设，可得左边 = \( (x + y)^{k+\!+} = (x + y)^{k} \times (x + y) = (\sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^jy^{k - j}) \times (x + y) = ((\sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^jy^{k - j}) \times x) + ((\sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^jy^{k - j}) \times y) = (\sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^{j + 1}y^{k - j}) + (\sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^{j}y^{k - j + 1}) \)，我们想合并\( \sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^{j + 1}y^{k - j} \) 和\( \sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^{j}y^{k - j + 1} \)，但对比每两个对应的项，我们会发现前者的项总是多了一个\( x \)，而后者的项总是多了一个\( y \)，导致无法合并，为了合并，我们可以将前者的每项向后移动一个位置（使用引理7.1.4的命题2），这样两者中间的项就能合并了，但前者的最后一项和后者的第一项得拿出来单独处理（读者如果知道什么是错位相加的话，应该可以识别出我们接下来要做的就是错位相加），具体的：前者 = \( \sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^{j + 1}y^{k - j} = \sum\limits_{j = 1}^{k+\!+} \dfrac{k!}{(j- 1)!(k - j + 1)!}x^{j}y^{k - j + 1} = (\sum\limits_{j = 1}^{k} \dfrac{k!}{(j - 1)!(k - j + 1)!}x^{j}y^{k - j + 1}) + x^{k + 1} \)，后者 = \( \sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^{j}y^{k - j + 1} = \sum\limits_{j = 0}^{k} \dfrac{k!}{j!(k - j)!}x^{j}y^{k - j + 1} = (\sum\limits_{j = 1}^{k} \dfrac{k!}{j!(k - j)!}x^{j}y^{k - j + 1}) + y^{k + 1} \)，接着前面的推，有左边 = \( (x + y)^{k+\!+} = \) 前者 \( + \) 后者 \( = ((\sum\limits_{j = 1}^{k} \dfrac{k!}{j!(k - j)!}x^{j}y^{k - j + 1}) + (\sum\limits_{j = 1}^{k} \dfrac{k!}{(j - 1)!(k - j + 1)!}x^{j}y^{k - j + 1})) + (x^{k + 1} + y^{k + 1}) = (\sum\limits_{j = 1}^{k} (\dfrac{k!}{j!(k - j)!} + \dfrac{k!}{(j - 1)!(k - j + 1)!})x^{j}y^{k - j + 1}) + (x^{k + 1} + y^{k + 1}) = (\sum\limits_{j = 1}^{k} \dfrac{k!((k - j + 1) + j)}{j!(k - j + 1)!}x^{j}y^{k - j + 1}) + (x^{k + 1} + y^{k + 1}) = (\sum\limits_{j = 1}^{k} \dfrac{(k + 1)!}{j!(k - j + 1)!}x^{j}y^{k - j + 1}) + (x^{k + 1} + y^{k + 1}) = (\sum\limits_{j = 1}^{k} \dfrac{(k + 1)!}{j!(k - j + 1)!}x^{j}y^{k - j + 1}) + (\dfrac{(k + 1)!}{(k + 1)!(k - (k + 1) + 1)!}x^{k + 1}y^{k - (k + 1) + 1}) + (\dfrac{(k + 1)!}{0!(k - 0 + 1)!}x^{0}y^{k - 0 + 1}) = \sum\limits_{j = 0}^{k+\!+} \dfrac{(k+\!+)!}{j!((k+\!+) - j)!}x^{j}y^{(k+\!+) - j} \)，即\( n = k+\!+ \)时成立。

至此，归纳完毕，命题成立。

证毕。

练习7.1.5

题目：

Let \( X \) be a finite set, let \( m \) be an integer, and for each \( x \in X \) let \( (a_n(x))_{n = m}^{\infty} \) be a convergent sequence of real numbers. Show that the sequence \( (\sum\limits_{x \in X} (a_n(x)))_{n = m}^{\infty} \) is convergent, and \( \lim\limits_{n \to \infty} \sum\limits_{x \in X} a_n(x) = \sum\limits_{x \in X} \lim\limits_{n \to \infty} a_n(x) \).

证明：

记\( X \)的基数为\( c \)，对\( c \)进行数学归纳。

当\( c = 0 \)时，\( (\sum\limits_{x \in X} (a_n(x)))_{n = m}^{\infty} = (0)_{n = m}^{\infty} \)为常数序列，收敛到\( 0 \)，而左边 = \( \lim\limits_{n \to \infty} \sum\limits_{x \in X} a_n(x) = \lim\limits_{n \to \infty} 0 = 0 \)，右边 = \( \sum\limits_{x \in X} \lim\limits_{n \to \infty} a_n(x) = 0 \)，可得左边 = 右边。 综上，\( c = 0 \)时成立。

归纳假设当\( c = k \)时成立，当\( c = k+\!+ \)时，因为\( X \)非空，故\( \exists x_0 \in X \)，令\( Y = X \setminus \{ x_0 \} \)，可得\( Y \)的基数为\( k \)且存在双射函数\( g: \{ i \in \mathbf{N} : 1 \leq i \leq k \} \to Y \)。

先证明\( (\sum\limits_{x \in X} (a_n(x)))_{n = m}^{\infty} \)收敛：根据归纳假设，有\( (\sum\limits_{x \in Y} (a_n(x)))_{n = m}^{\infty} = (\sum\limits_{i = 1}^{k} (a_n(g(i))))_{n = m}^{\infty} \)为收敛序列，构造函数\( g_1: \{ i \in \mathbf{N} : 1 \leq i \leq k+\!+ \} \to X \)， \( \forall 1 \leq i \leq k+\!+ \)，如果\( i < k+\!+ \)，则\( g_1(i) = g(i) \)，如果\( i = k+\!+ \)，则\( g_1(i) = x_0 \)，易证\( g_1 \)为双射函数，进而可得\( (\sum\limits_{x \in X} (a_n(x)))_{n = m}^{\infty} = (\sum\limits_{i = 1}^{k+\!+} (a_n(g_1(i))))_{n = m}^{\infty} = ((\sum\limits_{i = 1}^{k} (a_n(g_1(i)))) + a_n(g_1(k+\!+)))_{n = m}^{\infty} = ((\sum\limits_{i = 1}^{k} (a_n(g_1(i)))) + a_n(x_0))_{n = m}^{\infty} = ((\sum\limits_{i = 1}^{k} (a_n(g(i)))) + a_n(x_0))_{n = m}^{\infty} = (\sum\limits_{i = 1}^{k} (a_n(g(i))))_{n = m}^{\infty} + (a_n(x_0))_{n = m}^{\infty} \)，因为\( (\sum\limits_{i = 1}^{k} (a_n(g(i))))_{n = m}^{\infty} \)和\( (a_n(x_0))_{n = m}^{\infty} \) 均为收敛序列（前者是根据归纳假设来的，后者是题目给的），根据极限的加法运算法则，有\( \sum\limits_{x \in X} (a_n(x)))_{n = m}^{\infty} \)收敛。

最后证明 \( \lim\limits_{n \to \infty} \sum\limits_{x \in X} a_n(x) = \sum\limits_{x \in X} \lim\limits_{n \to \infty} a_n(x) \)：根据归纳假设，有\( \lim\limits_{n \to \infty} \sum\limits_{x \in Y} a_n(x) = \sum\limits_{x \in Y} \lim\limits_{n \to \infty} a_n(x) \)，即\( \lim\limits_{n \to \infty} (\sum\limits_{i = 1}^{k} (a_n(g(i)))) = \sum\limits_{i = 1}^{k} (\lim\limits_{n \to \infty} a_n(g(i))) \)，此时可得左边 = \( \lim\limits_{n \to \infty} \sum\limits_{x \in X} a_n(x) = \lim\limits_{n \to \infty} (\sum\limits_{i = 1}^{k+\!+} (a_n(g_1(i)))) = \lim\limits_{n \to \infty} ((\sum\limits_{i = 1}^{k} (a_n(g_1(i)))) + a_n(g_1(k+\!+))) = \lim\limits_{n \to \infty} ((\sum\limits_{i = 1}^{k} (a_n(g(i)))) + a_n(g_1(k+\!+))) = \lim\limits_{n \to \infty} (\sum\limits_{i = 1}^{k} (a_n(g(i)))) + \lim\limits_{n \to \infty} a_n(g_1(k+\!+)) = (\sum\limits_{i = 1}^{k} (\lim\limits_{n \to \infty} a_n(g(i)))) + \lim\limits_{n \to \infty} a_n(g_1(k+\!+)) = (\sum\limits_{i = 1}^{k} (\lim\limits_{n \to \infty} a_n(g_1(i)))) + \lim\limits_{n \to \infty} a_n(g_1(k+\!+)) = \sum\limits_{i = 1}^{k+\!+} (\lim\limits_{n \to \infty} a_n(g_1(i))) = \sum\limits_{x \in X} \lim\limits_{n \to \infty} a_n(x) \) = 右边。

综上，\( n = k+\!+ \)时成立，至此，归纳完毕，命题成立。

证毕。

章节7.2

练习7.2.1

题目：

Is the series \( \sum_{n = 1}^{\infty} (-1)^n \) convergent or divergent? Justify your answer. Can you now resolve the difficulty in Example 1.2.2?

解答：

因为序列\( ((-1)^n)_{n = 1}^{\infty} \)非柯西序列，故发散，根据推论7.2.6，可知\( \sum_{n = 1}^{\infty} (-1)^n \)发散。

为什么Example 1.2.2中后两个级数\( S \)似乎有多个不一样的“和”

Example 1.2.2中给出了三个级数\( S \)，分别是：

1. \( S = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dots \)
2. \( S = 1 + 2 + 4 + 8 + 16 + \dots \)
3. \( S = 1 - 1 + 1 - 1 + 1 - 1 + \dots \)

第一个是收敛的级数，它的和是有定义的，且根据极限的唯一性，它的和是唯一的。后两个级数 对应的序列 发散，故根据推论7.2.6，可得后两个级数发散，故它们的和是未定义的，此时利用各种技巧计算它们的“和”是“非法”的操作。

练习7.2.2

题目：

Prove Proposition 7.2.5. (Hint: use Proposition 6.1.12 and Theorem 6.4.18.)

Proposition 7.2.5的内容：

Let \( \sum_{n = m}^{\infty} a_n \) be a formal series of real numbers. Then \( \sum_{n = m}^{\infty} a_n \) converges if and only if, for every real number \( \epsilon > 0 \), there exists an integer \( N \geq m \) such that \( |\sum_{n = p}^{q} a_n| \leq \epsilon \) for all \( p, q \geq N \).

证明：

因为\( \sum_{n = m}^{\infty} a_n \)收敛，所以它的部分和序列\( (S_K)_{n = m}^{\infty} \)收敛，进而有\( (S_K)_{n = m}^{\infty} \)为柯西序列，因此\( \forall \epsilon > 0, \exists N \geq m, \forall q, p - 1 \geq N, |S_q - S_{p - 1}| \leq \epsilon \)，不妨设\( q \geq p - 1 \)（如果不是的话，直接交换两个变量就行），此时可得\( |S_q - S_{p - 1}| = |(\sum_{n = m}^{q} a_n) - (\sum_{n = m}^{p - 1} a_n)| = |\sum_{n = p}^{q} a_n| \leq \epsilon \)，这里因为\( p - 1 \geq N \)，故\( p \)也\( \geq N \)。简而言之，\( \forall \epsilon > 0, \exists N \geq m, \forall q, p \geq N, |\sum_{n = p}^{q} a_n| \leq \epsilon \)。

证毕。

练习7.2.3

题目：

Use Proposition 7.2.5 to prove Corollary 7.2.6.

Corollary 7.2.6的内容：

(Zero test). Let \( \sum_{n = m}^{\infty} a_n \) be a convergent series of real numbers. Then we must have \( \lim_{n \to \infty} a_n = 0 \). To put this another way, if \( \lim_{n \to \infty} a_n \) is non-zero or divergent, then the series \( \sum_{n = m}^{\infty} a_n \) is divergent.

证明：

因为\( \sum_{n = m}^{\infty} a_n \)收敛，根据定理7.2.5，有\( \forall \epsilon > 0, \exists N \geq m, \forall p, q \geq N, |\sum_{n = p}^{q} a_n| \leq \epsilon \)，特别的，有\( \forall k \geq N, |\sum_{n = k}^{k} a_n| = |a_k| = |a_k - 0| \leq \epsilon \)。简而言之，\( \forall \epsilon > 0, \exists N \geq m, \forall k \geq N, |a_k - 0| \leq \epsilon \)，即\( \lim_{n \to \infty} a_n = 0 \)。

证毕。

练习7.2.4

题目：

Prove Proposition 7.2.9. (Hint: use Proposition 7.2.5 and Proposition 7.1.4(5).)

Proposition 7.2.9的内容：

(Absolute convergence test). Let \( \sum_{n = m}^{\infty} a_n \) be a formal series of real numbers. If this series is absolutely convergent, then it is also conditionally convergent. Furthermore, in this case we have the triangle inequality \( |\sum_{n = m}^{\infty} a_n| \leq \sum_{n = m}^{\infty} |a_n| \).

证明：

因为\( \sum_{n = m}^{\infty} a_n \)绝对收敛，即\( \sum_{n = m}^{\infty} |a_n| \)收敛，根据定理7.2.5，可得\( \forall \epsilon > 0, \exists N \geq m, \forall p, q \geq m, |\sum_{n = p}^{q} |a_n|| \leq \epsilon \)，由定理7.1.4的命题5，可得\( |\sum_{n = p}^{q} a_n| \leq |\sum_{n = p}^{q} |a_n|| \leq \epsilon \)，综上，\( \sum_{n = m}^{\infty} a_n \)也收敛。

下面证明不等式\( |\sum_{n = m}^{\infty} a_n| \leq \sum_{n = m}^{\infty} |a_n| \)：

注：下面用到了定理7.1.4中的部分命题，引理6.4.13（因为收敛，故刚好收敛于上下极限，故可以用该引理）以及极限运算法则等，为了写得比较流畅点，没有明确指出来，读者自己注意下。

因为\( \forall n \geq m, -|a_n| \leq a_n \leq |a_n| \)，可得\( \forall N \geq m, \sum_{n = m}^{N} (-|a_n|) = -(\sum_{n = m}^{N} |a_n|) \leq \sum_{n = m}^{N} a_n \leq \sum_{n = m}^{N} |a_n| \) （注：这里类似\( \sum_{n = m}^{N} a_n \)的表达式是部分和的表达式，我没写成\( S_N \)的形式），故\( \lim\limits_{N \to \infty} (-(\sum_{n = m}^{N} |a_n|)) = -(\lim\limits_{N \to \infty} (\sum_{n = m}^{N} |a_n|)) \leq \lim\limits_{N \to \infty} (\sum_{n = m}^{N} a_n) \leq \lim\limits_{N \to \infty} (\sum_{n = m}^{N} |a_n|) \)，即\( -(\sum_{n = m}^{\infty} |a_n|) \leq \sum_{n = m}^{\infty} a_n \leq \sum_{n = m}^{\infty} |a_n| \)，这等价于\( |\sum_{n = m}^{\infty} a_n| \leq \sum_{n = m}^{\infty} |a_n| \)。

证毕。

练习7.2.5

题目：

Prove Proposition 7.2.14. (Hint: use Theorem 6.1.19.)

Proposition 7.2.14的内容：

(Series laws).

1. If \( \sum_{n = m}^{\infty} a_n \) is a series of real numbers converging to \( x \), and \( \sum_{n = m}^{\infty} b_n \) is a series of real numbers converging to \( y \), then \( \sum_{n = m}^{\infty} (a_n + b_n) \) is also a convergent series, and converges to \( x + y \). In particular, we have \( \sum_{n = m}^{\infty} (a_n + b_n) = \sum_{n = m}^{\infty} a_n + \sum_{n = m}^{\infty} b_n \).
2. If \( \sum_{n = m}^{\infty} a_n \) is a series of real numbers converging to \( x \), and \( c \) is a real number, then \( \sum_{n = m}^{\infty} (ca_n) \) is also a convergent series, and converges to \( cx \). In particular, we have \( \sum_{n = m}^{\infty} (ca_n) = c\sum_{n = m}^{\infty} a_n \).
3. Let \( \sum_{n = m}^{\infty} a_n \) be a series of real numbers, and let \( k \geq 0 \) be an integer. If one of the two series \( \sum_{n = m}^{\infty} a_n \) and \( \sum_{n = m + k}^{\infty} a_n \) are convergent, then the other one is also, and we have the identity \( \sum_{n = m}^{\infty} a_n = \sum_{n = m}^{m + k - 1} a_n + \sum_{n = m + k}^{\infty} a_n \).
4. Let \( \sum_{n = m}^{\infty} a_n \) be a series of real numbers converging to \( x \), and let \( k \) be an integer. Then \( \sum_{n = m + k}^{\infty} a_{n - k} \) also converges to \( x \).

• 命题1

  证明：

  因为\( \sum_{n = m}^{\infty} a_n \)收敛到\( x \)，故\( \lim\limits_{N \to \infty} (\sum_{n = m}^{N} a_n) = x \)，因为\( \sum_{n = m}^{\infty} b_n \)收敛到\( y \)，故\( \lim\limits_{N \to \infty} (\sum_{n = m}^{N} b_n) = y \)，根据极限的加法运算法则以及引理7.1.4的命题3，有\( x + y = \lim\limits_{N \to \infty} (\sum_{n = m}^{N} a_n) + \lim\limits_{N \to \infty} (\sum_{n = m}^{N} b_n) = \lim\limits_{N \to \infty} ((\sum_{n = m}^{N} a_n) + (\sum_{n = m}^{N} b_n)) = \lim\limits_{N \to \infty} (\sum_{n = m}^{N} (a_n + b_n)) \)，即\( \sum_{n = m}^{\infty} (a_n + b_n) = x + y = \sum_{n = m}^{\infty} a_n + \sum_{n = m}^{\infty} b_n \)。

  证毕。

• 命题2

  证明：

  因为\( \sum_{n = m}^{\infty} a_n \)收敛到\( x \)，故\( \lim\limits_{N \to \infty} (\sum_{n = m}^{N} a_n) = x \)，根据极限的数乘运算法则以及引理7.1.4的命题4，有\( cx = \lim\limits_{N \to \infty} c(\sum_{n = m}^{N} a_n) = \lim\limits_{N \to \infty} (\sum_{n = m}^{N} (ca_n)) \)，即\( \sum_{n = m}^{\infty} (ca_n) = cx = c\sum_{n = m}^{\infty} a_n \)。

  证毕。

• 命题3

  证明：

  如果\( \sum_{n = m}^{\infty} a_n \)收敛，则根据推论7.2.6，有\( \forall \epsilon > 0, \exists N \geq m, \forall p, q \geq N, |\sum_{n = p}^{q} a_n| \leq \epsilon \)，此时取\( N_0 = N + k \)，有\( N_0 \geq m + k \)，而\( \forall p, q \geq N_0 \)，有\( p, q \geq N \)，从而有\( |\sum_{n = p}^{q} a_n| \leq \epsilon \)。简而言之，\( \forall \epsilon > 0, \exists N_0 \geq m + k, \forall p, q \geq N, |\sum_{n = p}^{q} a_n| \leq \epsilon \)，即\( \sum_{n = m + k}^{\infty} a_n \)收敛。

  反之，如果\( \sum_{n = m + k}^{\infty} a_n \)收敛，则根据推论7.2.6，有\( \forall \epsilon > 0, \exists N \geq m + k, \forall p, q \geq N, |\sum_{n = p}^{q} a_n| \leq \epsilon \)，既然\( N \geq m + k \)，可得\( N \geq m \)，即\( \exists N \geq m, \forall p, q \geq N, |\sum_{n = p}^{q} a_n| \leq \epsilon \)，综上可得，\( \sum_{n = m}^{\infty} a_n \)收敛。

  根据引理7.1.4的命题1以及极限的加法运算法则，可得 \( \sum_{n = m}^{\infty} a_n = \lim\limits_{N \to \infty} (\sum_{n = m}^{N} a_n) = \lim\limits_{N \to \infty} ((\sum_{n = m}^{m + k - 1} a_n) + (\sum_{n = m + k}^{N} a_n)) = \lim\limits_{N \to \infty} (\sum_{n = m}^{m + k - 1} a_n) + \lim\limits_{N \to \infty} (\sum_{n = m + k}^{N} a_n) = \sum_{n = m}^{m + k - 1} a_n + \lim\limits_{N \to \infty} (\sum_{n = m + k}^{N} a_n) = \sum_{n = m}^{m + k - 1} a_n + \sum_{n = m + k}^{\infty} a_n \) （注：\( \lim\limits_{N \to \infty} (\sum_{n = m}^{m + k - 1} a_n) = \sum_{n = m}^{m + k - 1} a_n \)，因为这是常数序列）。

  证毕。

• 命题4

  证明：

  因为\( \sum_{n = m}^{\infty} a_n \)收敛到\( x \)，故\( \lim\limits_{N \to \infty} (\sum_{n = m}^{N} a_n) = x \)，可得\( \forall \epsilon > 0 \)，有\( \epsilon / 2 > 0 \) 且\(\exists K_0 \geq m, \forall N \geq K_0, |(\sum_{n = m}^{N} a_n) - x| \leq \epsilon / 2 \)，再根据推论7.2.6，有\( \exists K_1 \geq m, \forall p, q \geq K_1, |\sum_{n = p}^{q} a_n| \leq \epsilon / 2 \)，取\( K = \max(K_0, K_1) \)，可得\( \forall N \geq K \)，有：

  1. \( |(\sum_{n = m}^{N} a_n) - x| \leq \epsilon / 2 \)
  2. 因为\( N, N + k \geq K \)，故\( |\sum_{n = N}^{N + k} a_n| \leq \epsilon / 2 \)

  此时，根据引理7.1.4的命题1、命题2以及绝对值不等式，可得： \( |(\sum_{n = m}^{N} a_n) - x| = |(\sum_{n = m + k}^{N + k} a_{n - k}) - x| = |((\sum_{n = m + k}^{N} a_{n - k}) + (\sum_{n = N + 1}^{N + k} a_{n - k})) - x| \leq |(\sum_{n = m + k}^{N} a_{n - k}) - x| + |\sum_{n = N + 1}^{N + k} a_{n - k}| \leq (\epsilon / 2) + (\epsilon / 2) = \epsilon \)，特别的，由\( |(\sum_{n = m + k}^{N} a_{n - k}) - x| + |\sum_{n = N + 1}^{N + k} a_{n - k}| \leq \epsilon \)，可得\( |(\sum_{n = m + k}^{N} a_{n - k}) - x| \leq \epsilon \)。

  综上可得， \( \forall \epsilon > 0, \exists K \geq m, \forall N \geq K, |(\sum_{n = m + k}^{N} a_{n - k}) - x| \leq \epsilon \)，即\( \lim\limits_{N \to \infty} (\sum_{n = m + k}^{N} a_{n - k}) = x \)，也就是\( \sum_{n = m + k}^{\infty} a_{n - k} \)收敛到\( x \)。

  证毕。

练习7.2.6

题目：

Prove Lemma 7.2.15. (Hint: First work out what the partial \( \sum_{n = 0}^{N} (a_n - a_{n + 1}) \) should be, and prove your assertion using induction.) How does the proposition change if we assume that \( (a_n)_{n = 0}^{\infty} \) does not converge to zero, but instead converges to some other real number \( L \)?

Lemma 7.2.15的内容：

(Telescoping series). Let \( (a_n)_{n = 0}^{\infty} \) be a sequence of real a numbers which converge to \( 0 \), i.e., \( \lim_{n \to \infty} a_n = 0 \). Then the series \( \sum_{n = 0}^{\infty} (a_n - a_{n + 1}) \) converges to \( a_0 \).

证明：

我们首先证明，\( \forall N \geq m, \sum_{n = 0}^{N} (a_n - a_{n + 1}) = a_0 - a_{N + 1} \)，使用数学归纳法证明。

当\( N = 0 \)时，\( \sum_{n = 0}^{N} (a_n - a_{n + 1}) = \sum_{n = 0}^{0} (a_n - a_{n + 1}) = a_0 - a_{0 + 1} = a_0 - a_{N + 1} \)，即\( N = 0 \)时成立。

归纳假设当\( N = K \)时成立，当\( N = K+\!+ \)时，根据归纳假设，有 \( \sum_{n = 0}^{N} (a_n - a_{n + 1}) = \sum_{n = 0}^{K+\!+} (a_n - a_{n + 1}) = (\sum_{n = 0}^{K} (a_n - a_{n + 1})) + (a_{K+\!+} - a_{(K+\!+) + 1}) = (a_0 - a_{K + 1}) + (a_{K+\!+} - a_{(K+\!+) + 1}) = a_0 - a_{(K+\!+) + 1} \)，即\( N = K+\!+ \)时成立。

至此，归纳完毕，有\( \forall N \geq m, \sum_{n = 0}^{N} (a_n - a_{n + 1}) = a_0 - a_{N + 1} \)。

根据极限的加法和数乘运算法则（或者说减法运算法则）以及\( \lim\limits_{N \to \infty} a_{N + 1} = 0 \)，有\( \sum_{n = 0}^{\infty} (a_n - a_{n + 1}) = \lim\limits_{N \to \infty} (\sum_{n = 0}^{N} (a_n - a_{n + 1})) = \lim\limits_{N \to \infty} (a_0 - a_{N + 1}) = \lim\limits_{N \to \infty} a_0 - \lim\limits_{N \to \infty} a_{N + 1} = \lim\limits_{N \to \infty} a_0 - 0 = a_0 \)，即\( \sum_{n = 0}^{\infty} (a_n - a_{n + 1}) = a_0 \)。

证毕。

如果\( (a_n)_{n = 0}^{\infty} \)不收敛到0，而是收敛到某个其他的实数\( L \)会怎样？

如果\( (a_n)_{n = 0}^{\infty} \)收敛到某个其他的实数\( L \)，则前面的证明中，最后的步骤变成： \( \sum_{n = 0}^{\infty} (a_n - a_{n + 1}) = \lim\limits_{N \to \infty} (\sum_{n = 0}^{N} (a_n - a_{n + 1})) = \lim\limits_{N \to \infty} (a_0 - a_{N + 1}) = \lim\limits_{N \to \infty} a_0 - \lim\limits_{N \to \infty} a_{N + 1} = a_0 - L \)，也就是结论变成\( \sum_{n = 0}^{\infty} (a_n - a_{n + 1}) = a_0 - L \)而已，仍然是收敛级数。

章节7.3

练习7.3.1

题目：

Use Proposition 7.3.1 to prove Corollary 7.3.2.

Corollary 7.3.2的内容：

(Comparison test). Let \( \sum_{n = m}^{\infty} a_n \) and \( \sum_{n = m}^{\infty} b_n \) be two formal series of real numbers, and suppose that \( |a_n| \leq b_n \) all \( n \geq m \). Then if \( \sum_{n = m}^{\infty} b_n \) is convergent, then \( \sum_{n = m}^{\infty} a_n \) is absolutely convergent, and in fact \( |\sum_{n = m}^{\infty} a_n| \leq \sum_{n = m}^{\infty} |a_n| \leq \sum_{n = m}^{\infty} b_n \).

证明：

如果\( \sum_{n = m}^{\infty} b_n \)收敛，则根据定理7.3.1，可得\( \exists M \in \mathbf{R}, \forall N \geq m, \sum_{n = m}^{N} b_n \leq M \)，又\( |a_n| \leq b_n \)，可得\( \sum_{n = m}^{N} |a_n| \leq \sum_{n = m}^{N} b_n \leq M \)，再根据定理7.3.1，有\( \sum_{n = m}^{\infty} |a_n| \)收敛。

根据定理7.2.9，可得\( \sum_{n = m}^{\infty} a_n \)也收敛且 \( |\sum_{n = m}^{\infty} a_n| \leq \sum_{n = m}^{\infty} |a_n| \)。因为\( \forall N \geq m, \sum_{n = m}^{N} |a_n| \leq \sum_{n = m}^{N} b_n \)，根据引理6.4.13（因为收敛，故刚好收敛于上下极限，故可以用该引理），可得\( \lim_{N \to \infty} \sum_{n = m}^{N} |a_n| \leq \lim_{N \to \infty} \sum_{n = m}^{N} b_n \)，即\( \sum_{n = m}^{\infty} |a_n| \leq \sum_{n = m}^{\infty} b_n \)。综上，有\( |\sum_{n = m}^{\infty} a_n| \leq \sum_{n = m}^{\infty} |a_n| \leq \sum_{n = m}^{\infty} b_n \)。

证毕。

练习7.3.2

题目：

Prove Lemma 7.3.3. (Hint: for the first part, use the zero test. For the second part, first use induction to establish the geometric series formula \( \sum_{n = 0}^{N} x^n = (1 - x^{N + 1}) / (1 - x) \) and then apply Lemma 6.5.2.

Lemma 7.3.3的内容：

(Geometric series). Let \( x \) be a real number. If \( |x| \geq 1 \) then the series \( \sum_{n = 0}^{\infty} x^n \) is divergent. If however \( |x| < 1 \), then the series is absolutely convergent and \( \sum_{n = 0}^{\infty} x_n = 1 / (1 - x) \).

证明：

根据引理6.5.2，如果\( |x| > 1 \)或\( x = -1 \)，则\( (x^n)_{n = 0}^{\infty} \)发散，如果\( x = 1 \)，则\( \lim_{n \to \infty} x^n = 1 \)，总而言之，当\( |x| \geq 1 \)时， \( (x^n)_{n = 0}^{\infty} \)均不收敛于\( 0 \)，根据推论7.2.6，有\( \sum_{n = 0}^{\infty} x_n \)发散。

现证明\( \forall N \geq 0, \sum_{n = 0}^{N} x^n = (1 - x^{N + 1}) / (1 - x) \)：

当\( N = 0 \)时，左边 = \( \sum_{n = 0}^{N} x^n = \sum_{n = 0}^{0} x^n = x^0 = 1 \)，右边 = \( (1 - x^{N + 1}) / (1 - x) = (1 - x^{1}) / (1 - x) = 1 \)，左边 = 右边，即\( N = 0 \)时成立。

归纳假设当\( N = K \)时成立，当\( N = K + 1 \)时，根据归纳假设，可得\( \sum_{n = 0}^{N} x^n = \sum_{n = 0}^{K + 1} x^n = (\sum_{n = 0}^{K} x^n) + x^{K + 1} = ((1 - x^{K + 1}) / (1 - x)) + x^{K + 1} = ((1 - x^{K + 1}) + ((1 - x)x^{K + 1})) / (1 - x) = (1 - x^{K + 2}) / (1 - x) \)，即\( N = K + 1 \)时成立。

至此，归纳完毕，有\( \forall N \geq 0, \sum_{n = 0}^{N} x^n = (1 - x^{N + 1}) / (1 - x) \)。

下面继续证明命题的后半部分，当\( |x| < 1 \)时，根据引理6.5.2，可得\( \lim_{n \to \infty}^{\infty} x^n = 0 \)，根据极限的乘法运算法则，可得\( \lim_{N \to \infty}^{\infty} x^{N + 1} = (\lim_{N \to \infty}^{\infty} x^N)(\lim_{N \to \infty}^{\infty} x) = 0x = 0 \)，再根据极限的运算法则可得， \( \lim_{N \to \infty} (\sum_{n = 0}^{N} x^n) = \lim_{N \to \infty} ((1 - x^{N + 1}) / (1 - x)) = (1 - 0) / (1 - x) = 1 / (1 - x) \)，即\( \sum_{n = 0}^{\infty} x_n = 1 / (1 - x) \)。

证毕。

练习7.3.3

题目：

Let \( \sum_{n = 0}^{\infty} a_n \) be an absolutely convergent series of real numbers such that \( \sum_{n = 0}^{\infty} |a_n| = 0 \). Show that \( a_n = 0 \) for every natural number n.

证明：

假设\( \exist n_0 \geq 0, a_n \neq 0 \)，则\( |a_n| > 0 \)，进而可得\( |a_n| = \sum_{n = 0}^{n_0} |a_n| > 0 \)，此时根据定理7.2.14的命题3，有\( \sum_{n = 0}^{\infty} |a_n| = \sum_{n = 0}^{n_0} |a_n| + \sum_{n = n_0 + 1}^{\infty} |a_n| > 0 \)，这和\( \sum_{n = 0}^{\infty} |a_n| = 0 \)矛盾，故假设不成立，有\( \forall n \geq 0, a_n = 0 \)。

证毕。

章节7.4

练习7.4.1

题目：

Let \( \sum_{n = 0}^{\infty} a_n \) be an absolutely convergent series of real numbers. Let \( f: \mathbf{N} \to \mathbf{N} \) be an increasing function (i.e., \( f(n + 1) > f(n) \) for all \( n \in \mathbf{N} \)). Show that \( \sum_{n = 0}^{\infty} a_{f(n)} \) is also an absolutely convergent series. (Hint: try to compare each partial sum of \( \sum_{n = 0}^{\infty} a_{f(n)} \) with a (slightly different) partial sum of \( \sum_{n = 0}^{\infty} a_n \).)

证明：

因为\( \sum_{n = 0}^{\infty} a_n \)绝对收敛，故\( \sum_{n = 0}^{\infty} |a_n| \)收敛，根据定理7.2.5，可得\( \forall \epsilon > 0, \exists N \geq 0, \forall p, q \geq N, |\sum_{n = p}^{q} |a_n|| \leq \epsilon \)。因为\( f \)为递增函数，易证\( \forall n \geq 0, f(n) \geq n \)。 \( \forall p, q \geq N \)，不妨设\( p \leq q \)（如果不满足的话，则交换两个变量），记\( X = \{ n \in \mathbf{N} : p \leq n \leq q \} \)，\( Y = \{ n \in \mathbf{N} : p \leq n \leq f(q) \} \)，因为\( p, f(q) \geq N \)，可得\( |\sum_{n \in Y} |a_n|| = |\sum_{n = p}^{f(q)} |a_n|| \leq \epsilon \)，而\( \forall x \in f(X) \)，有\( f(p) \leq x \leq f(q) \)，又\( f(p) \geq p \)，从而有\( p \leq x \leq f(q) \)，可得\( x \in Y \)，此时可知\( f(X) \subseteq Y \)，进而可得\( |\sum_{n = p}^{q} |a_{f(n)}|| = |\sum_{n \in f(X)} |a_n|| \leq |\sum_{n \in Y} |a_n|| \leq \epsilon \)。简而言之，\( \forall \epsilon > 0, \exists N \geq 0, \forall p, q \geq N, |\sum_{n = p}^{q} |a_{f(n)}|| \leq \epsilon \)，再次根据定理7.2.5，可得\( \sum_{n = 0}^{\infty} |a_{f(n)|} \)收敛，即\( \sum_{n = 0}^{\infty} a_{f(n)} \)绝对收敛。

证毕。

章节7.5

练习7.5.1

题目：

Prove the first inequality in Lemma 7.5.2.

Lemma 7.5.2的内容：

Let \( (c_n)_{n = m}^{\infty} \) be a sequence of positive numbers. Then we have \( \lim\inf_{n \to \infty} \dfrac{c_{n + 1}}{c_n} \leq \lim\inf_{n \to \infty} c_n^{1 / n} \leq \lim\sup_{n \to \infty} c_n^{1 / n} \leq \lim\sup_{n \to \infty} \dfrac{c_{n + 1}}{c_n} \).

证明：

记\( L = \lim\inf_{n \to \infty} \dfrac{c_{n + 1}}{c_n} \)，因为\( (\dfrac{c_{n + 1}}{c_n})_{n = m}^{\infty} \)序列中每项都是正数，可得\( L \geq 0 \)， \( \forall 0 < \epsilon < L \)，有\( 0 < L - \epsilon < L \)，根据定理6.4.12的命题1，可得\( \exists N \geq m, \forall n \geq N, L - \epsilon < \dfrac{c_{n + 1}}{c_n} \)，可得\( (L - \epsilon)c_n < c_{n + 1} \)，使用数学归纳法，易证\( \forall n \geq N, (L - \epsilon)^{n - N} < c_n \)，记\( A = (L - \epsilon)^{-N} \)，不等式可写成\( A(L - \epsilon)^n < c_n \)，进而可得\( A^{1 / n}(L - \epsilon) < c_n^{1 / n} \)，根据引理6.4.13，可得\( \lim\sup_{n \to \infty} A^{1 / n}(L - \epsilon) \leq \lim\sup_{n \to \infty} c_n^{1 / n} \)，针对该不等式的左边，因为\( A > 0 \)，根据引理6.5.3，可得\( \lim_{n \to \infty} A^{1 / n} = 1 \)，故\( \lim_{n \to \infty} A^{1 / n}(L - \epsilon) = 1 \times (\lim_{n \to \infty} (L - \epsilon)) = L - \epsilon \)，回代到不等式，可得\( L - \epsilon \leq \lim\sup_{n \to \infty} c_n^{1 / n} \)，该不等式针对任意\( 0 < \epsilon < L \)都成立，可得\( L \leq \lim\sup_{n \to \infty} c_n^{1 / n} \)，理由：假设\( L > \lim\sup_{n \to \infty} c_n^{1 / n} \)，则取\( \epsilon_0 = \min((L - (\lim\sup_{n \to \infty} c_n^{1 / n})) / 2, L / 2) \)，可得\( 0 < \epsilon_0 < L \)且\( L - \epsilon_0 > \lim\sup_{n \to \infty} c_n^{1 / n} \)，然而这和前面得到的结论\( L - \epsilon_0 \leq \lim\sup_{n \to \infty} c_n^{1 / n} \)矛盾，故假设不成立，有\( L \leq \lim\sup_{n \to \infty} c_n^{1 / n} \)，即\( \lim\inf_{n \to \infty} \dfrac{c_{n + 1}}{c_n} \leq \lim\sup_{n \to \infty} c_n^{1 / n} \)。

证毕。

练习7.5.2

题目：

Let \( x \) be a real number with \( |x| < 1 \), and \( q \) be a real number. Show that the series \( \sum_{n = 1}^{\infty} n^qx^n \) is absolutely convergent, and that \( \lim_{n \to \infty} n^qx^n = 0 \).

证明：

当\( x = 0 \)时，\( \forall n \geq 1, n^qx^n = n^q \times 0 = 0 \)，可得\( \sum_{n = 1}^{\infty} |n^qx^n| = \lim_{N \to \infty} \sum_{n = 1}^{N} |n^qx^n| = \lim_{N \to \infty} \sum_{n = 1}^{N} |0| = \lim_{N \to \infty} 0 = 0 \)，即\( \sum_{n = 1}^{\infty} n^qx^n \)绝对收敛。

当\( x \neq 0 \)时，\( \forall n \geq 1 \)，有\( n^qx^n \neq 0 \)，故可以使用推论7.5.3（比例测试）。此时\( \dfrac{|(n + 1)^qx^{n + 1}|}{|n^qx^n|} = |\dfrac{(n + 1)^qx^{n + 1}}{n^qx^n}| = |(\dfrac{n + 1}{n})^qx| \)，下面证明\( \lim_{n \to \infty} (\dfrac{n + 1}{n})^q = \lim_{n \to \infty} (1 + \dfrac{1}{n})^q = 1 \) （注：我们没有证明过极限的指数法则，故不能直接得出该结论），根据极限的加法运算法则，可得\( \lim_{n \to \infty} (1 + \dfrac{1}{n}) = 1 + 0 = 1 \)，故\( \forall \epsilon > 0 \)，有\( \epsilon^{1 / q} > 0 \)，进而有\( \exists N \geq 1, \forall n \geq N, |1 + \dfrac{1}{n} - 1| = |\dfrac{1}{n}| \leq \epsilon^{1 / q} \)，可得\( |\dfrac{1}{n}|^q = |(\dfrac{1}{n})^q| = |(1 + (\dfrac{1}{n})^q) - 1| \leq (\epsilon^{1 / q})^q = \epsilon \)，简而言之，\( \forall \epsilon > 0, \exists N \geq 1, \forall n \geq N, |(1 + (\dfrac{1}{n})^q) - 1| \leq \epsilon \)，即\( \lim_{n \to \infty} (\dfrac{n + 1}{n})^q = \lim_{n \to \infty} (1 + \dfrac{1}{n})^q = 1 \)。根据极限的乘法运算法则，可得\( \lim_{n \to \infty} \dfrac{|(n + 1)^qx^{n + 1}|}{|n^qx^n|} = \lim_{n \to \infty} |\dfrac{(n + 1)^qx^{n + 1}}{n^qx^n}| = \lim_{n \to \infty} |(\dfrac{n + 1}{n})^qx| = \lim_{n \to \infty} |(\dfrac{n + 1}{n})^q||x| = (\lim_{n \to \infty} |(\dfrac{n + 1}{n})^q|)(\lim_{n \to \infty} |x|) = |x| < 1 \)，根据推论7.5.3，可得\( \sum_{n = 1}^{\infty} n^qx^n \)绝对收敛。

综上，当\( |x| < 1 \)时，\( \sum_{n = 1}^{\infty} n^qx^n \)绝对收敛，从而可得\( \sum_{n = 1}^{\infty} n^qx^n \)本身也收敛（条件收敛），此时根据定理7.5.4（零测试），可得\( \lim_{n \to \infty} n^qx^n = 0 \)。

证毕。

练习7.5.3

题目：

Give an example of a divergent series \( \sum_{n = 1}^{\infty} a_n \) of positive numbers \( a_n \) such that \( \lim_{n \to \infty} a_{n + 1} / a_n = \lim_{n \to \infty} a_n^{1 / n} = 1 \), and give an example of a convergent series \( \sum_{n = 1}^{\infty} b_n \) of positive numbers \( b_n \) such that \( \lim_{n \to \infty} b_{n + 1} / b_n = \lim_{n \to \infty} b_n^{1 / n} = 1 \). (Hint: use Corollary 7.3.7.) This shows that the ratio and root tests can be inconclusive even when the summands are positive and all the limits converge.

解答：

令\( q > 0 \)为任意有理数，针对级数\( \sum_{n = 1}^{\infty} (1 / n^q) \)，可得\( \forall n \geq 1, 1 / n^q \)均为正数，且\( \lim_{n \to \infty} ((1 / (n + 1)^q) / (1 / n^q)) = \lim_{n \to \infty} (n / (n + 1))^q = \lim_{n \to \infty} (((n + 1) / n)^{-1})^q = \lim_{n \to \infty} (((n + 1) / n)^{-q} = \lim_{n \to \infty} (((n + 1) / n)^q)^{-1} \)，在练习7.5.2中的证明中，我们证明了\( \lim_{n \to \infty} ((n + 1) / n)^q = 1 \)，根据极限的倒数运算法则，有\( \lim_{n \to \infty} ((1 / (n + 1)^q) / (1 / n^q)) = \lim_{n \to \infty} (((n + 1) / n)^q)^{-1} = 1 \)。

根据推论7.3.7，当\( q > 1 \)时，\( \sum_{n = 1}^{\infty} (1 / n^q) \)发散，当\( q \leq 1 \)时，\( \sum_{n = 1}^{\infty} (1 / n^q) \)收敛。令\( (a_n)_{n = 1}^{\infty} = (1 / n)_{n = 1}^{\infty} \)，可得\( \lim_{n \to \infty} (a_{n + 1} / a_n) = \lim_{n \to \infty} ((1 / (n + 1)) / (1 / n)) = 1 \) 且\( \sum_{n = 1}^{\infty} a_n \)收敛。令\( (b_n)_{n = 1}^{\infty} = (1 / n^2)_{n = 1}^{\infty} \)，可得\( \lim_{n \to \infty} (b_{n + 1} / b_n) = \lim_{n \to \infty} ((1 / (n + 1)^2) / (1 / n^2)) = 1 \) 且\( \sum_{n = 1}^{\infty} b_n \)发散。