习题5.3

第 1 题

举出相应的例子.

(1) \(u_n > 0, \lim\limits_{n \to \infty}u_n = 0\), 但 \(\sum\limits_{n = 1}^{\infty}(-1)^{n}u_n\) 发散;

(2) \(u_n > 0\), \(\{u_n\}\) 单调递减, 但 \(\sum\limits_{n = 1}^{\infty}(-1)^{n}u_n\) 发散.

解

(1) 构造 \(u_n = \begin{cases}\dfrac{2}{k}, & n = 2k-1, \\ \dfrac{1}{k}, & n = 2k,\end{cases}\) 则 \(\lim\limits_{n \to \infty}u_n = 0\). 但 \(\sum\limits_{n = 1}^{\infty}(-1)^{n}u_n = \sum\limits_{k = 1}^{\infty}\dfrac{1}{k}\) 发散.

(2) 构造 \(u_n = 1 + \dfrac{1}{n}\), 则 \(\sum\limits_{n = 1}^{\infty}(-1)^{n}u_n = \sum\limits_{n = 1}^{\infty}(-1)^n + \sum\limits_{n = 1}^{\infty}\dfrac{(-1)^n}{n}\). 由于前者发散, 后者收敛, 故两者之和发散.

第 3 题

若级数 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛, 且 \(\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} = 1\), 能否断定级数 \(\sum\limits_{n = 1}^{\infty}v_n\) 收敛?

解

不可以. 构造 \(u_n = \dfrac{(-1)^n}{\sqrt{n}}, v_n = \dfrac{(-1)^n}{\sqrt{n}} + \dfrac{1}{n}\), 则满足

\[ \begin{aligned} \lim\limits_{n \to \infty}\dfrac{u_n}{v_n} & = \lim\limits_{n \to \infty}\dfrac{\frac{(-1)^n}{\sqrt{n}}}{\frac{(-1)^n}{\sqrt{n}} + \frac{1}{n}} \\ & = \lim\limits_{n \to \infty} \dfrac{\sqrt{n}}{\sqrt{n} + (-1)^n} \\ & = 1 \end{aligned} \]

但显然 \(\sum\limits_{n = 1}^{\infty}v_n\) 不收敛.

第 4 题

判断下列级数绝对收敛、条件收敛还是发散.

(1) \(\sum\limits_{n = 1}^{\infty}\dfrac{(-1)^n}{\sqrt{n + 1}}\);

(3) \(\sum\limits_{n = 1}^{\infty}(-1)^{n}\dfrac{n}{n + 1}\);

(5) \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{2^n}\sin\dfrac{n\pi}{4}\);

(7) \(\sum\limits_{n = 1}^{\infty}(-1)^{n}\dfrac{2^{n^2}}{n!}\);

(9) \(\sum\limits_{n = 1}^{\infty}(-1)^n(\sqrt{n + 1} - \sqrt{n})\);

(11) \(\sum\limits_{n = 2}^{\infty}\dfrac{(-1)^n}{\sqrt{n + (-1)^n}}\);

(13) \(\dfrac{1}{\sqrt{2} - 1} + \dfrac{1}{\sqrt{2} + 1} + \dfrac{1}{\sqrt{3} - 1} + \dfrac{1}{\sqrt{3} + 1} + \cdots + \dfrac{1}{\sqrt{n} - 1} + \dfrac{1}{\sqrt{n} + 1} + \cdots\);

(14) \(1 - \ln 2 + \dfrac{1}{2} - \dfrac{3}{2} + \dfrac{1}{n} - \ln\dfrac{n + 1}{n} + \cdots\).

解

(1) 由于 \(\sum\limits_{n = 1}^{\infty}\dfrac{(-1)^n}{\sqrt{n + 1}}\) 为 Leibniz 级数, 故该级数收敛. 而

\[ \sum\limits_{n = 1}^{\infty}\left|(-1)^{n}\dfrac{1}{\sqrt{n + 1}}\right| = \sum\limits_{n = 1}^{\infty}\dfrac{1}{\sqrt{n + 1}} \]

发散. 故原级数条件收敛.

(3) 由于 \(\sum\limits_{n = 1}^{\infty}(-1)^{n}\dfrac{n}{n + 1} = \sum\limits_{n = 1}^{\infty}(-1)^n + \sum\limits_{n = 1}^{\infty}\dfrac{(-1)^{n - 1}}{n + 1}\), 且前者发散, 后者收敛, 故原级数发散.

(5) 由于 \(\left|\dfrac{1}{2^n}\sin\dfrac{n\pi}{4}\right|\le \dfrac{1}{2^n}\) 且 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{2^n}\) 收敛, 故原级数绝对收敛.

(7) 由于

\[ \begin{aligned} \sum\limits_{n = 1}^{\infty}(-1)^{n}\dfrac{2^{n^2}}{n!} & = \sum\limits_{k = 1}^{\infty}\left(-\dfrac{2^{(2k-1)^2}}{(2k - 1)!} + \dfrac{2^{(2k)^2}}{(2k)!}\right) \\ & = \sum\limits_{k = 1}^{\infty}\dfrac{2^{(2k - 1)^2}}{(2k)!}\cdot(2^{4k - 1} - 2k) \end{aligned} \]

显然 \(\lim\limits_{k \to \infty}\dfrac{2^{(2k - 1)^2}}{(2k)!}\cdot(2^{4k - 1} - 2k) \neq 0\). 故该级数发散.

(9) 由于 \(\sum\limits_{n = 1}^{\infty}(-1)^n(\sqrt{n + 1} - \sqrt{n})\) 为 Leibniz 级数, 故该级数收敛. 而

\[ \begin{aligned} \sum\limits_{n = 1}^{\infty}\left|(-1)^n(\sqrt{n + 1} - \sqrt{n})\right| & = \sum\limits_{n = 1}^{\infty}\left(\sqrt{n + 1} - \sqrt{n}\right) \\ & = \lim\limits_{n \to \infty}\sqrt{n + 1} - 1 \end{aligned} \]

故该级数的通项加上绝对值后发散. 综上所述, 该级数条件收敛.

(11) 由于

\[ \begin{aligned} \sum\limits_{n = 2}^{\infty}\dfrac{(-1)^n}{\sqrt{n + (-1)^n}} & = \sum\limits_{k = 1}^{\infty}\left(\dfrac{(-1)^{2k}}{\sqrt{2k + (-1)^{2k}}} + \dfrac{(-1)^{2k + 1}}{\sqrt{2k + 1 + (-1)^{2k + 1}}}\right) \\ & = \sum\limits_{k = 1}^{\infty}\left(\dfrac{1}{\sqrt{2k + 1}} - \dfrac{1}{\sqrt{2k}}\right) \\ & = \sum\limits_{n = 2}^{\infty}\dfrac{(-1)^{n - 1}}{\sqrt{n}} \end{aligned} \]

为 Leibniz 级数, 因此该级数收敛. 而

\[ \begin{aligned} \sum\limits_{n = 2}^{\infty}\left|\dfrac{(-1)^n}{\sqrt{n + (-1)^n}}\right| & = \sum\limits_{n = 1}^{\infty}\dfrac{1}{\sqrt{n + (-1)^{n}}} \\ & \ge \sum\limits_{n = 1}^{\infty}\dfrac{1}{\sqrt{n + 1}} \\ \end{aligned} \]

故该级数的通项加上绝对值后发散. 综上所述, 该级数条件收敛.

(13)

\[ \begin{aligned} & \quad \dfrac{1}{\sqrt{2} - 1} + \dfrac{1}{\sqrt{2} + 1} + \dfrac{1}{\sqrt{3} - 1} + \dfrac{1}{\sqrt{3} + 1} + \cdots + \dfrac{1}{\sqrt{n} - 1} + \dfrac{1}{\sqrt{n} + 1} + \cdots \\ & = \sum\limits_{n = 2}^{\infty}\left(\dfrac{1}{\sqrt{n} - 1} + \dfrac{1}{\sqrt{n} + 1}\right) \\ & = \sum\limits_{n = 2}^{\infty}\dfrac{2\sqrt{n}}{n - 1} \\ & \ge \sum\limits_{n = 2}^{\infty}\dfrac{2}{n - 1} \end{aligned} \]

故原级数发散.

(14)

\[ \begin{aligned} & \quad 1 - \ln 2 + \dfrac{1}{2} - \dfrac{3}{2} + \dfrac{1}{n} - \ln\dfrac{n + 1}{n} + \cdots \\ & = \sum\limits_{n = 1}^{\infty}\left(\dfrac{1}{n} - \ln\dfrac{n + 1}{n}\right) \\ & = \lim\limits_{n \to \infty}\left(\sum\limits_{k = 1}^{n}\dfrac{1}{k} - \ln(n + 1)\right) \\ & = \lim\limits_{n \to \infty}\left(\ln n + \gamma + \epsilon_n - \ln(n + 1)\right) \\ & = \gamma \end{aligned} \]

其中 \(\lim\limits_{n \to \infty}\epsilon_n = 0, \gamma = 0.577216\cdots\). 故原级数绝对收敛.

第 6 题

级数 \(\sum\limits_{n = 1}^{\infty}a_n^2, \sum\limits_{n = 1}^{\infty}b_n^2\) 收敛, 证明: \(\sum\limits_{n = 1}^{\infty}(a_n + b_n)^2\) 收敛, \(\sum\limits_{n = 1}^{\infty}\dfrac{a_n}{n}\) 绝对收敛.

证明

由于 \((a_n + b_n)^2 \le 2(a_n^2 + b_n^2)\), 且 \(\sum\limits_{n = 1}^{\infty}2(a_n^2 + b_n^2)\) 收敛, 故由比较判别法知 \(\sum\limits_{n = 1}^{\infty}(a_n + b_n)^2\) 收敛.

不妨设 \(a_n > 0\). 由于 \(\dfrac{a_n}{n} \le a_n^2 + \dfrac{1}{4n^2}\), 且 \(\sum\limits_{n = 1}^{\infty}\left(a_n^2 + \dfrac{1}{4n^2}\right) = \sum\limits_{n = 1}^{\infty}a_n^2 + \sum\limits_{n = 1}^{\infty}\dfrac{1}{4n^2}\) 收敛, 故由比较判别法知 \(\sum\limits_{n = 1}^{\infty}\dfrac{a_n}{n}\) 绝对收敛.

第 7 题

若级数 \(\sum\limits_{n = 1}^{\infty}a_n, \sum\limits_{n = 1}^{\infty}b_n\) 均收敛, 且 \(a_n \le c_n \le b_n\), 证明: \(\sum\limits_{n = 1}^{\infty}c_n\) 收敛.

证明

由于 \(0 \le c_n - a_n \le b_n - a_n\), 且 \(\sum\limits_{n = 1}^{\infty}(b_n - a_n)\) 收敛, 故由比较判别法知 \(\sum\limits_{n = 1}^{\infty}(c_n - a_n)\) 收敛. 又因为 \(\sum\limits_{n = 1}^{\infty}a_n\) 收敛, 故 \(\sum\limits_{n = 1}^{\infty}c_n\) 收敛.

第 9 题

设正项数列 \(\{u_n\}\) 单调减少, 且级数 \(\sum\limits_{n = 1}^{\infty}(-1)^nu_n\) 发散, 证明: 级数 \(\sum\limits_{n = 1}^{\infty}\left(\dfrac{1}{1 + u_n}\right)^n\) 收敛.

证明

由于正项数列 \(\{u_n\}\) 单调减少, 故 \(\lim\limits_{n \to \infty}u_n \ge 0\). 假设 \(\lim\limits_{n \to \infty}u_n = 0\), 则级数 \(\sum\limits_{n = 1}^{\infty}(-1)^nu_n\) 收敛, 矛盾. 故 \(\lim\limits_{n \to \infty}u_n > 0\), 不妨设为 \(c\). 由于

\[ \lim\limits_{n \to \infty}\sqrt[n]{\left(\dfrac{1}{1 + u_n}\right)^n} = \lim\limits_{n \to \infty}\dfrac{1}{1 + u_n} = \dfrac{1}{1 + c} < 1 \]

由根值判别法知级数 \(\sum\limits_{n = 1}^{\infty}\left(\dfrac{1}{1 + u_n}\right)^n\) 收敛.