级数的敛散性

Abstract（摘要）

本节题目主要涉及数项级数的定义，Cauchy准则，以及数项级数的一些性质

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Table of Contents（目录）

1. 1. 问题陈述（Problem Statement）
2. 2. 预备知识（Preliminaries）
3. 3. 主要结果与证明（Main Result & Proof）
4. 4. 注记与讨论（Remarks & Discussion）
5. 5. 相关拓展（Related Extensions）
6. 6. 参考文献（References）

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1. 问题陈述（Problem Statement）

1. 证明下列级数收敛,并求其和：

(1) \(\displaystyle \frac{1}{1 \cdot 6}+\frac{1}{6 \cdot 11}+\frac{1}{11 \cdot 16}+\dots+\frac{1}{(5n-4)(5n+1)}+\dots ;\)

(2) \(\displaystyle \left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}\right)+\dots+\left(\frac{1}{2^n}+\frac{1}{3^n}\right)+\dots ;\)

(3) \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} ;\)

(4) \(\displaystyle \sum_{n=1}^{\infty} (\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}) ;\)

(5) \(\displaystyle \sum_{n=1}^{\infty} \frac{2n-1}{2^n}.\)

2.证明:若级数 \(\sum u_n\) 发散,\(c \neq 0\),则 \(\sum c u_n\) 也发散.

3.设级数 \(\sum u_n\) 与 \(\sum v_n\) 都发散,试问 \(\sum (u_n+v_n)\) 一定发散吗? 又若 \(u_n\) 与 \(v_n(n = 1,2,\dots)\)都是非负数,则能得出什么结论?

4.证明:若数列 \(\{a_n\}\) 收敛于 \(a\),则级数 \(\displaystyle \sum_{n=1}^{\infty} \left(a_n-a_{n+1}\right) = a_1-a.\)

5.证明: 若数列 \(\{b_n\}\) 有 \(\displaystyle \lim_{n \to \infty} b_n = \infty\), 则
(1) 级数 \(\displaystyle \sum (b_{n+1}-b_n)\) 发散；
(2) 当 \(b_n \neq 0\) 时,级数 \(\displaystyle \sum \left(\frac{1}{b_n}-\frac{1}{b_{n+1}}\right) = \frac{1}{b_1}.\)
6. 应用第4,5题的结果求下列级数的和：
(1) \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)}\)；

(2) \(\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1} \frac{2n+1}{n(n+1)}\)；

(3) \(\displaystyle \sum_{n=1}^{\infty} \frac{2n+1}{(n^2 + 1)\left((n+1)^2 + 1\right)}\)。
7. 应用柯西准则判别下列级数的敛散性：
(1) \(\displaystyle \sum \frac{\sin 2^n}{2^n}\)；
(2) \(\displaystyle \sum \frac{(-1)^{n-1} n^2}{2n^2 + 1}\)；
(3) \(\displaystyle \sum \frac{(-1)^n}{n}\)；
(4) \(\displaystyle \sum \frac{1}{\sqrt{n+n^2}}\)。
8.证明级数 \(\sum u_n\) 收敛的充要条件是:任给正数 \(\varepsilon\),存在某正整数 \(N\),对一切 \(n > N\) 总有

\[|u_N+u_{N+1}+\dots +u_n| < \varepsilon. \]

9.举例说明:若级数 \(\sum u_n\) 对每个固定的 \(p\) 满足条件

\[\lim_{n \to \infty} \left(u_{n+1}+u_{n+2}+\dots +u_{n+p}\right) = 0, \]

此级数仍可能不收敛。

10.设级数 \(\sum u_n\) 满足:加括号后级数 \(\displaystyle \sum_{k=1}^{\infty} \left(u_{n_k + 1}+u_{n_k + 2}+\dots +u_{n_{k+1}}\right)\) 收敛\((n_1=0)\),且在同一括号中的 \(u_{n_k+1},u_{n_k+2},\dots ,u_{n_{k+1}}\) 符号相同,证明:\(\sum u_n\) 亦收敛。

2. 预备知识（Preliminaries）

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3. 主要结果与证明（Main Result & Proof）

1.\(\frac{1}{1 \cdot 6}+\frac{1}{6 \cdot 11}+\frac{1}{11 \cdot 16}+\dots+\frac{1}{(5n-4)(5n+1)}+\dots\)

解:

\[\begin{align*} S_n &= \sum_{k=1}^{n} \frac{1}{(5k-4)(5k+1)} = \sum_{k=1}^{n} \frac{1}{5}\left(\frac{1}{5k-4}-\frac{1}{5k+1}\right) \\ &= \frac{1}{5}\left(1-\frac{1}{6}\right)+\frac{1}{5}\left(\frac{1}{6}-\frac{1}{11}\right)+\dots+\frac{1}{5}\left(\frac{1}{5n-4}-\frac{1}{5n+1}\right) \\ &= \frac{1}{5}\left(1-\frac{1}{5n+1}\right) \end{align*} \]

\[\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{5}\left(1-\frac{1}{5n+1}\right) = \frac{1}{5} \]

故 \(\sum \frac{1}{(5n-4)(5n+1)}\) 收敛，其和为 \(\frac{1}{5}\).\(\Box\)

2.\(\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}\right)+\dots+\left(\frac{1}{2^n}+\frac{1}{3^n}\right)+\dots\)

解:

\[\begin{align*} S_n &= \sum_{k=1}^{n} \left(\frac{1}{2^k}+\frac{1}{3^k}\right) = \sum_{k=1}^{n} \frac{1}{2^k} + \sum_{k=1}^{n} \frac{1}{3^k} \\ &= \frac{\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^n\right]}{1-\frac{1}{2}} + \frac{\frac{1}{3}\left[1-\left(\frac{1}{3}\right)^n\right]}{1-\frac{1}{3}} \\ &= 1-\left(\frac{1}{2}\right)^n + \frac{1}{2}\left(1-\left(\frac{1}{3}\right)^n\right) \\ &= \frac{3}{2}-\left(\frac{1}{2}\right)^n-\frac{1}{2}\left(\frac{1}{3}\right)^n \to \frac{3}{2} \quad (n \to \infty) \end{align*} \]

\(\Box\)
3.\(\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}\)

解:

\[S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)} \]

设

\[u_k = \frac{1}{k(k+1)(k+2)} = \frac{A}{k} + \frac{Bk+C}{(k+1)(k+2)}(待定系数法) \]

故

\[A(k+1)(k+2) + k(Bk+C) = 1 \]

解得

\[\begin{cases} A = \displaystyle \frac{1}{2} \\[5pt] B = -\displaystyle \frac{1}{2} \\[5pt] C = -\displaystyle \frac{3}{2} \end{cases} \]

重复上述步骤，有

\[u_k = \frac{\frac{1}{2}}{k} + \frac{-\frac{1}{2}k-\frac{3}{2}}{(k+1)(k+2)} = \frac{\frac{1}{2}}{k} + \frac{D}{k+1} + \frac{E}{k+2}(待定系数法) \]

故

\[D(k+2) + E(k+1) = -\frac{1}{2}k-\frac{3}{2} \]

解得

\[\begin{cases} D = -1 \\[5pt] E = \displaystyle \frac{1}{2} \end{cases} \]

因此

\[\begin{align*} S_n &= \sum_{k=1}^{n} u_k = \sum_{k=1}^{n} \left( \frac{\frac{1}{2}}{k} + \frac{-1}{k+1} + \frac{\frac{1}{2}}{k+2} \right) \\ &= \frac{1}{4} + \frac{1}{2}\cdot\frac{1}{n+2} - \frac{1}{2}\cdot\frac{1}{n+1} \to \frac{1}{4} \quad (n\to\infty) \end{align*} \]

故 \(\displaystyle \sum \frac{1}{n(n+1)(n+2)}\) 收敛，其和为 \(\displaystyle \frac{1}{4}\).\(\Box\)
4.\(\sum_{n=1}^{\infty} \left( \sqrt{n+2}-2\sqrt{n+1}+\sqrt{n} \right)\)

解:

\[\begin{align*} S_n &= \sum_{k=1}^{n} \left( \sqrt{k+2}-2\sqrt{k+1}+\sqrt{k} \right) \\ &= \sum_{k=1}^{n} \left( \sqrt{k+2}-\sqrt{k+1} \right) - \sum_{k=1}^{n} \left( \sqrt{k+1}-\sqrt{k} \right) \\ &= \left( \sqrt{n+2}-\sqrt{2} \right) - \left( \sqrt{n+1}-\sqrt{1} \right) \\ &= \frac{(n+2)-(n+1)}{\sqrt{n+2}+\sqrt{n+1}} +1-\sqrt{2} \to 1-\sqrt{2} \quad (n\to\infty) \end{align*} \]

故 \(\displaystyle \sum_{n=1}^{\infty} \left( \sqrt{n+2}-2\sqrt{n+1}+\sqrt{n} \right)\) 收敛，其和为 \({1-\sqrt{2}}\).\(\Box\)

5.\(\sum_{n=1}^{\infty} \frac{2n-1}{2^n}\)

解:

\[S_n = \sum_{k=1}^{n} \frac{2k-1}{2^k} \quad ① \]

\[\frac{1}{2}S_n = \sum_{k=1}^{n} \frac{2k-1}{2^{k+1}} \quad ② \]

\(①-②\)得

\[\begin{align*} S_n-\frac{1}{2}S_n &= \frac{1}{2}+\frac{3}{2^2}-\frac{1}{2^2}+\frac{5}{2^3}-\frac{3}{2^3}+\dots+\frac{2n-1}{2^n}-\frac{2n-3}{2^n}-\frac{2n-1}{2^{n+1}} \\ &= \frac{1}{2}+\sum_{k=1}^{n-1} \frac{2}{2^{k+1}} -\frac{2n-1}{2^{n+1}} \\ &= \frac{1}{2}-\frac{2n-1}{2^{n+1}} + 2\cdot \frac{\frac{1}{2^2}\left(1-(\frac{1}{2})^{n-1}\right)}{1-\frac{1}{2}} \\ &= \frac{1}{2}-\frac{2n-1}{2^{n+1}} +1-\left(\frac{1}{2}\right)^{n-1} \to \frac{3}{2} \quad (n\to\infty) \end{align*} \]

故 \(\displaystyle \sum_{n=1}^{\infty} \frac{2n-1}{2^n}\) 收敛，其和为 \(\displaystyle \lim_{n \to \infty} 2 \cdot \lim_{n \to \infty} \frac{3}{2} = 3\).\(\Box\)

2. 证明: 若级数 \(\sum u_n\) 发散，\(C \neq 0\)，则 \(\sum Cu_n\) 也发散

(证法一)
证明:(反证法)若 \(\sum Cu_n\) 收敛，则由 \(C \neq 0\)可知

\[\frac{1}{C}\sum Cu_n = \sum \frac{1}{C}(Cu_n) = \sum u_n \]

收敛，与 \(\sum u_n\) 发散矛盾，故 \(\sum Cu_n\) 发散.\(\Box\)

(证法二)
证明:$\exists \varepsilon_0>0,\ \forall N>0,\ \exists\ n_0>N,\ p_0>0,\ $ 有

\[\left| u_{n_0+1}+u_{n_0+2}+\dots+u_{n_0+p_0} \right| \ge \varepsilon_0 \]

由 \(C \neq 0\) 可知 \(|C|>0\)

因此

\[\begin{align*} \left| Cu_{n_0+1}+Cu_{n_0+2}+\dots+Cu_{n_0+p_0} \right| &= |C| \cdot \left| u_{n_0+1}+u_{n_0+2}+\dots+u_{n_0+p_0} \right| \ge |C| \varepsilon_0 \end{align*} \]

故 \(\sum Cu_n\) 发散.\(\Box\)

3. 设级数 \(\sum u_n\) 与 \(\sum v_n\) 都发散，试问 \(\sum (u_n+v_n)\) 一定发散吗？又若 \(u_n\) 与 \(v_n\ (n=1,2,\dots)\) 都是非负数，则能得出什么结论？

(1)解:不一定.如级数

\[\sum u_n = \sum n \]

\[\sum v_n = \sum (-n) \]

(2)证明:(反证法)若 \(\sum (u_n+v_n)\) 收敛，则对 \(\forall \varepsilon>0,\ \exists N>0\)，当 \(n>N\)，对 \(\forall p\) 有

\[\left| (u_{n+1}+v_{n+1}) + \dots + (u_{n+p}+v_{n+p}) \right| < \varepsilon \]

因为\(u_n\) 与 \(v_n\ (n=1,2,\dots)\) 都是非负数，所以

\[u_{n+1} + \dots + u_{n+p} < \varepsilon \]

\[v_{n+1} + \dots + v_{n+p} < \varepsilon \]

故 $ \sum u_n,\ \sum v_n$ 收敛，这与级数 \(\sum u_n\) 与 \(\sum v_n\) 都发散矛盾，因此 \(\sum (u_n+v_n)\) 发散.\(\Box\)

4. 证明：若数列 \(\{a_n\}\) 收敛于 \(a\)，则级数\( \sum_{n=1}^{\infty} \left( a_n - a_{n+1} \right) = a_1 - a \)

\[S_n = \sum_{k=1}^{n} \left( a_k - a_{k+1} \right) = a_1 - a_2 + a_2 - a_3 + \dots + a_n - a_{n+1} = a_1 - a_{n+1} \]

由 $$a_n \to a\ (n\to\infty)$$

得 $$S_n \to a_1 - a\ (n\to\infty)$$
\(\Box\)
5. 证明：若数列 \(\{b_n\}\) 有 \(\displaystyle \lim_{n \to \infty} b_n = \infty\)，则

(1) 级数 \(\sum \left( b_{n+1}-b_n \right)\) 发散；

(2) 当 \(b_n \neq 0\) 时，级数 \(\sum \left( \frac{1}{b_n} - \frac{1}{b_{n+1}} \right) = \frac{1}{b_1}\).
证明:(1)级数 \(\sum \left( b_{n+1}-b_n \right)\)的部分和为

\[S_n = \sum_{k=1}^{n} \left( b_{k+1}-b_k \right) = b_2 - b_1 + b_3 - b_2 + \dots + b_{n+1} - b_n = b_{n+1} - b_1 \]

由 \(\{b_{n+1}\}\)发散可知\(\{S_n\}\)发散.\(\Box\)

故 \(\sum \left( b_{n+1}-b_n \right)\) 发散.\(\Box\)

(2)部分和

\[S_n = \sum_{k=1}^{n} \left( \frac{1}{b_k} - \frac{1}{b_{k+1}} \right) = \frac{1}{b_1} - \frac{1}{b_2} + \frac{1}{b_2} - \frac{1}{b_3} + \dots + \frac{1}{b_n} - \frac{1}{b_{n+1}} = \frac{1}{b_1} - \frac{1}{b_{n+1}} \]

由 \(b_n \neq 0\)，且 \(\displaystyle \lim_{n \to \infty} b_n = \infty\) 得

\[\lim_{n \to \infty} S_n = \frac{1}{b_1} \]

故级数 \(\sum \left( \frac{1}{b_n} - \frac{1}{b_{n+1}} \right) = \frac{1}{b_1}\).\(\Box\)

6. (1)\( \sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)}; \)

解:部分和

\[S_n = \sum_{k=1}^{n} \frac{1}{(a+k-1)(a+k)} \]

设

\[u_k = \frac{1}{(a+k-1)(a+k)} = \frac{A}{a+k-1} + \frac{B}{a+k}(待定系数法) \]

故

\[A(a+k) + B(a+k-1) = 1 \]

解得

\[\begin{cases} A = 1 \\ B = -1 \end{cases} \]

因此

\[\begin{align*} S_n &= \sum_{k=1}^{n} u_k = \sum_{k=1}^{n} \left( \frac{1}{a+k-1} - \frac{1}{a+k} \right) \\ &= \frac{1}{a} - \frac{1}{a+1} + \frac{1}{a+1} - \frac{1}{a+2} + \dots + \frac{1}{a+n-1} - \frac{1}{a+n} \\ &= \frac{1}{a} - \frac{1}{a+n} \to \frac{1}{a} \quad (n\to\infty) \end{align*} \]

即得级数

\[\sum \frac{1}{(a+n-1)(a+n)} = \frac{1}{a} \]

\(\Box\)
(2)\( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{2n+1}{n(n+1)}; \)

解:部分和

\[S_n = \sum_{k=1}^{n} (-1)^{k+1} \frac{2k+1}{k(k+1)} = \sum_{k=1}^{n} (-1)^{k+1} \left( \frac{1}{k}+\frac{1}{k+1} \right) \]

\[= \left(1+\frac{1}{2}\right) - \left(\frac{1}{2}+\frac{1}{3}\right) + \left(\frac{1}{3}+\frac{1}{4}\right) - \left(\frac{1}{4}+\frac{1}{5}\right) + \dots + (-1)^n \left(\frac{1}{n-1}+\frac{1}{n}\right) + (-1)^{n+1}\left(\frac{1}{n}+\frac{1}{n+1}\right) \]

可以看到这里的一般项的符号受\(n\)的奇偶性的影响，由此想到数列极限与子列极限的关系
由

\[S_{2n} = \sum_{k=1}^{2n} (-1)^{k+1} \left( \frac{1}{k}+\frac{1}{k+1} \right) = 1 - \frac{1}{2n+1} \to 1\quad (n\to\infty) \]

及

\[S_{2n-1} = \sum_{k=1}^{2n-1} (-1)^{k+1} \left( \frac{1}{k}+\frac{1}{k+1} \right) = 1 + \frac{1}{2n} \to 1\quad (n\to\infty) \]

可知

\[\lim_{n \to \infty} S_n = \lim_{n \to \infty} S_{2n} = \lim_{n \to \infty} S_{2n-1} = 1 \]

因此级数

\[\sum (-1)^{n+1} \frac{2n+1}{n(n+1)} = 1 \]

\(\Box\)
(3)\( \sum_{n=1}^{\infty} \frac{2n+1}{\left(n^2+1\right)\left(\left(n+1\right)^2+1\right)}; \)

解:部分和

\[\begin{align*} S_n &= \sum_{k=1}^{n} \frac{2k+1}{(k^2+1)\left((k+1)^2+1\right)} \\ &= \sum_{k=1}^{n} \frac{2k+k^2+1-k^2}{(k^2+1)\left((k+1)^2+1\right)} \\ &= \sum_{k=1}^{n} \frac{(k+1)^2 - k^2}{(k^2+1)\left((k+1)^2+1\right)} \\ &= \sum_{k=1}^{n} \frac{\left[(k+1)^2+1\right] - (k^2+1)}{(k^2+1)\left((k+1)^2+1\right)} \\ &= \sum_{k=1}^{n} \left( \frac{1}{k^2+1} - \frac{1}{(k+1)^2+1} \right) \\ &= \frac{1}{2} - \frac{1}{(n+1)^2+1} \to \frac{1}{2} \quad (n\to\infty) \end{align*} \]

因此级数

\[\sum \frac{2n+1}{(n^2+1)\left((n+1)^2+1\right)} = \frac{1}{2} \]

\(\Box\)

7. 应用柯西准则判别下列级数的敛散性：

(1)

\[\sum \frac{\sin 2^n}{2^n}; \]

证明:\(\forall p \in \mathbb{N}\)，有

\[|S_{n+p}-S_n| = \left| \sum_{k=1}^{n+p} \frac{\sin 2^k}{2^k} - \sum_{k=1}^{n} \frac{\sin 2^k}{2^k} \right| = \left| \sum_{k=n+1}^{n+p} \frac{\sin 2^k}{2^k} \right| \leq \sum_{k=n+1}^{n+p} \frac{|\sin 2^k|}{2^k} \leq \sum_{k=n+1}^{n+p} \frac{1}{2^k} \]

\[= \frac{\frac{1}{2^{n+1}}\left(1-\left(\frac{1}{2}\right)^p\right)}{1-\frac{1}{2}} \leq \frac{\frac{1}{2^{n+1}}}{\frac{1}{2}} = \frac{1}{2^n} < \frac{1}{n} \]

对 \(\forall \varepsilon>0,\ \exists N=\left\lceil \frac{1}{\varepsilon} \right\rceil\)，当 \(n>N\) 时，\(\forall p \in \mathbb{N}\)，有

\[|S_{n+p}-S_n| < \frac{1}{n} < \varepsilon \]

由 Cauchy 准则，证得\(\displaystyle \sum \frac{\sin 2^n}{2^n}\) 收敛.\(\Box\)

(2)

\[\sum \frac{(-1)^{n-1} n^2}{2n^2+1}; \]

证明:有部分和

\[S_n = \sum_{k=1}^{n} \frac{(-1)^{k-1} k^2}{2k^2+1} \]

取 \(\varepsilon_0 = \frac{2}{5}\)，对 \(\forall N>0,\ \exists n_0>N\) 及 \(p=1\)，有

\[\begin{align*} |S_{n_0+1}-S_{n_0}| &= \left| \frac{(-1)^{n_0} (n_0+1)^2}{2(n_0+1)^2+1} \right| \\ &= \frac{(n_0+1)^2}{2(n_0+1)^2+1} \ge \frac{(n_0+1)^2}{2\left[(n_0+1)^2+1\right]} \\ &= \frac{(n_0+1)^2+1-1}{2\left[(n_0+1)^2+1\right]} \\ &= \frac{1}{2} - \frac{1}{2\left[(n_0+1)^2+1\right]} \\ &\ge \frac{1}{2}-\frac{1}{10} = \frac{2}{5} = \varepsilon_0 \end{align*} \]

由Cauchy准则的否定形式可知原级数发散.\(\Box\)
\((\)由 \(n_0>N\ge1\) 可知 \( 2\left[(n_0+1)^2+1\right] \ge 10 \)，即\( -\frac{1}{2\left[(n_0+1)^2+1\right]} \ge -\frac{1}{10}) \)

(3)

\[\sum \frac{(-1)^n}{n}^①; \]

证明:\(\forall\ p \in \mathbb{N}\)，有

\[\left| S_{n+p}-S_n \right| = \left| \frac{(-1)^{n+1}}{n+1} + \dots + \frac{(-1)^{n+p}}{n+p} \right| < \frac{1}{n+1} ^② \]

对\(\forall\ \varepsilon>0,\ \exists N = \left\lceil \frac{1}{\varepsilon} \right\rceil\)，当 \(n>N\) 且 \(\forall\ p \in \mathbb{N}\)，有

\[\left| S_{n+p}-S_n \right| \le \frac{1}{n+1} < \varepsilon \]

由Cauchy准则可知原级数收敛.\(\Box\)
注
①:其实由交错级数中的莱布尼茨判别法可以立即判定该级数收敛.

②:这里容易使用绝对值不等式进行估计，而使用之后就陷入调和级数的发散性证明中，因此使用上述估计，而涉及自然数的命题一般可用数学归纳法得到证明.
证明如下:设有数列 \(\{a_k\},\ k=1,2,\cdots\)

一般项

\[a_k = \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + \cdots + (-1)^k \frac{1}{n+k} \]

亦可表示为

\[a_k = \frac{1}{n} + \sum_{i=1}^{k} (-1)^i \frac{1}{n+i} \]

欲证：\(\forall\ k \in \mathbb{N}\)，有 \(a_k < \frac{1}{n}\)
(数学归纳法)
当 \(k=1\) 时

\[a_1 = \frac{1}{n} - \frac{1}{n+1} < \frac{1}{n} \]

命题成立

设对任意 \(k \in \mathbb{N}\)，有

\[a_k = \frac{1}{n} + \sum_{i=1}^{k} (-1)^i \frac{1}{n+i} < \frac{1}{n} \]

考察 \(k+1\) 的情形

\[a_{k+1} = \frac{1}{n} + \sum_{i=1}^{k+1} (-1)^i \frac{1}{n+i} \]

当 \(k\) 为偶数时

\[a_{k+1} = a_k + (-1)^{k+1} \frac{1}{n+k+1} < \frac{1}{n} - \frac{1}{n+k+1} < \frac{1}{n} \]

当 \(k\) 为奇数时

\[a_{k+1} = \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \cdots + (-1)^k \frac{1}{n+k} + \frac{1}{n+k+1} \]

且有

\[-\frac{1}{n+i} + \frac{1}{n+i+1} < 0,\quad i=1,2,\cdots,k \]

可得

\[a_{k+1} < \frac{1}{n} \]

由数学归纳法可知命题成立.\(\Box\)
(4)

\[\sum \frac{1}{\sqrt{n+n^2}}. \]

证明:\(\forall p \in \mathbb{N}\)，有

\[|S_{n+p}-S_n| = \frac{1}{\sqrt{n(n+1)}} + \dots + \frac{1}{\sqrt{(n+p)(n+p+1)}} \ge \frac{p}{\sqrt{(n+p)(n+p+1)}} \]

令 \(p=n\)

有

\[|S_{2n}-S_n| \ge \frac{n}{\sqrt{2n(2n+1)}} = \frac{n}{\sqrt{4n^2+2n}} \ge \frac{n}{\sqrt{8n^2}} = \frac{n}{2\sqrt{2}n} = \frac{\sqrt{2}}{4} \]

取 \(\varepsilon_0 = \frac{\sqrt{2}}{4} > 0\)，\(\forall\ N>0,\ \exists\ n>N\) 且 \(p=n\)，有

\[|S_{2n}-S_n| \ge \frac{\sqrt{2}}{4} = \varepsilon_0 \]

由Cauchy准则的否定形式可知原级数发散.

7. 证明级数 \(\sum u_n\) 收敛的充要条件是:任给正数 \(\varepsilon\),存在某正整数 \(N\),对一切 \(n>N\)
   总有

\[\left| u_{N}+u_{N+1}+\dots +u_{n} \right| < \varepsilon \]

证明:（⇐）
对 \(\forall \varepsilon>0,\ \exists N>0,\ \forall\ n>m>N\)

有

\[|u_N + u_{N+1} + \dots + u_n| < \varepsilon \]

\[|u_N + u_{N+1} + \dots + u_m| < \varepsilon \]

故

\[\begin{align*} |u_{m+1} + \dots + u_n| &= \left| \left(u_N+u_{N+1}+\dots+u_n\right)-\left(u_N+u_{N+1}+\dots+u_m\right) \right| \\ &\le \left|u_N+u_{N+1}+\dots+u_n\right|+\left|u_N+u_{N+1}+\dots+u_m\right| \\ &< 2\varepsilon \end{align*} \]

由Cauchy准则可知级数 \(\sum u_n\) 收敛.

（⇒）若级数 \(\sum u_n\) 收敛，由Cauchy准则可知，对 \(\forall \varepsilon>0,\ \exists N_1>0\)，当 \(n>m>N_1\) 时

有

\[|u_{m+1} + \dots + u_n| < \varepsilon \]

取 \(N>N_1+1\)，令 \(m=N-1>N_1\)，当 \(n>N\) 时

有

\[|u_N + u_{N+1} + \dots + u_n| < \varepsilon.\Box \]

9. 举例说明:若级数 \(\sum u_n\) 对每个固定的 \(p\) 满足条件

\[\lim_{n \to \infty} \left( u_{n+1}+u_{n+2}+\dots+u_{n+p} \right)=0, \]

此级数仍可能不收敛.

解:对于级数

\[\sum u_n = \sum \ln\left(1+\frac{1}{n}\right) \]

其部分和

\[S_n = \sum\limits_{k=1}^n \ln\frac{k+1}{k} = \ln(n+1) \]

对 \(\forall p \in N\)，有

\[\lim_{n \to \infty} \left(u_{n+1}+\dots+u_{n+p}\right) = \lim_{n \to \infty}\left(S_{n+p}-S_n\right) = \lim_{n \to \infty}\ln\left(1+\frac{p}{n+1}\right)=0 \]

而由 $$S_n =\ln(n+1)$$ 可知

\[\{S_n\} \]

发散，即 \(\sum \ln\left(1+\frac{1}{n}\right)\) 发散.

10. 设级数 \(\sum u_n\) 满足:加括号后级数 \(\sum\limits_{k=1}^{\infty} (u_{n_k + 1} + u_{n_k + 2} + \dots + u_{n_{k+1}})\) 收敛\((n_1 = 0)\),且在同一括号中的 \(u_{n_k + 1},u_{n_k + 2},\dots,u_{n_{k+1}}\) 符号相同 ,证明:\(\sum u_n\) 亦收敛.

证明: 因为级数 \(\sum\limits_{k=1}^{\infty} (u_{n_k+1}+u_{n_k+2}+\dots+u_{n_{k+1}})\) 收敛，则有

\[\lim_{n \to \infty} (u_{n_k+1}+u_{n_k+2}+\dots+u_{n_{k+1}})=0, \]

所以对 \(\forall n \in \mathbb{N}_+\), 总存在 \(k \in \mathbb{N}_+\), 使 \(n=n_k+j\ (1 \le j \le n_{k+1}-n_k)\) 时，有

\[\begin{align*} S_n &= \sum_{i=1}^n u_i \\ &= \sum_{i=1}^{k-1} \left( u_{n_i+1}+u_{n_i+2}+\dots+u_{n_{i+1}} \right) + \left( u_{n_k+1}+u_{n_k+2}+\dots+u_{n_k+j} \right) \\ &= S_{k-1}' + \left( u_{n_k+1}+u_{n_k+2}+\dots+u_{n_k+j} \right), \end{align*} \]

其中\(S_{k-1}'\)表示加括号级数的前\(k-1\)项之和. 当\(n\to\infty\)时,\(k-1\to \infty\),从而有

\[S= \lim_{n\to\infty} S_n= \lim_{n\to\infty} S_{k-1}' + \lim_{n\to\infty} \left( u_{n_k+1}+ u_{n_k+2}+ \dots + u_{n_k+j} \right)= \lim_{n\to\infty} S_{k-1}', \]

故 \(\sum u_n\) 收敛,其和不变.\(\Box\)

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4. 注记与讨论（Remarks & Discussion）

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5. 相关拓展（Related Extensions）

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6. 参考文献（References）

数学分析.下册/华东师范大学数学科学学院编--5 版.--北京:高等教育出版社，2019.5