习题5.2

第 1 题

判断下列级数的敛散性.

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

(3) \(\sum\limits_{n = 2}^{\infty}\dfrac{1}{\ln n}\);

(5) \(\sum\limits_{n = 1}^{\infty}\left(\dfrac{1 + n^2}{1 + n^3}\right)^2\);

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

解

(1) 由于 \(\lim\limits_{n \to \infty}\dfrac{u_{n + 1}}{u_n} = \dfrac{2n - 1}{2(2n + 1)} = \dfrac{1}{2} < 1\), 故由比值判别法知 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{(2n - 1)2^{n-1}}\) 收敛.

(3) 由于 \(\ln n \ge n - 1\), 因此 \(\sum\limits_{n = 2}^{\infty}\dfrac{1}{\ln n} \ge \sum\limits_{n = 1}^{\infty}\dfrac{1}{n}\). 显然 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{n}\) 发散. 故由比较判别法知 \(\sum\limits_{n = 2}^{\infty}\dfrac{1}{\ln n}\) 发散.

(5) 由于 \(\dfrac{1 + n^2}{1 + n^3} \le \dfrac{1}{n - 1}, \forall n \ge 2\), 所以 \(\sum\limits_{n = 1}^{\infty}\left(\dfrac{1 + n^2}{1 + n^3}\right)^2 \ge 1 + \sum\limits_{n = 2}^{\infty}\dfrac{1}{(n - 1)^2}\). 显然 \(\sum\limits_{n = 2}^{\infty}\dfrac{1}{(n - 1)^2}\) 收敛. 故由比较判别法知 \(\sum\limits_{n = 1}^{\infty}\left(\dfrac{1 + n^2}{1 + n^3}\right)^2\) 收敛.

(7) 由于 \(\lim\limits_{n \to \infty}n^{\frac{3}{2}} \cdot u_n = \lim\limits_{n \to \infty}n^{\frac{3}{2}} \cdot \dfrac{1}{\sqrt{n}}\ln\left(\dfrac{n + 1}{n - 1}\right) = \lim\limits_{n \to \infty}n\ln\left(\dfrac{n + 1}{n - 1}\right) = 2\), 故由推论 5.2.1 知 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{\sqrt{n}}\ln\left(\dfrac{n + 1}{n - 1}\right)\) 收敛.

第 2 题

判断下列级数的敛散性.

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

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

(5) \(\sum\limits_{n = 1}^{\infty}n^3\sin\dfrac{\pi}{3^n}\);

解

(1) 由于 \(\lim\limits_{n \to \infty}\dfrac{u_{n + 1}}{u_n} = \lim\limits_{n \to \infty}\dfrac{2}{2n + 1} = 0 < 1\), 故由比值判别法知 \(\sum\limits_{n = 1}^{\infty}\dfrac{2^n}{(2n - 1)!}\) 收敛.

(3) 由于 \(\lim\limits_{n \to \infty}\sqrt[n]{u_n} = \dfrac{3}{n}\sqrt[n]{n} = 0 < 1\), 故由根值判别法知 \(\sum\limits_{n = 1}^{\infty}\dfrac{3^nn}{n^n}\) 收敛.

(5) 由于 \(\sum\limits_{n = 1}^{\infty}n^3\sin\dfrac{\pi}{3^n} < \sum\limits_{n = 1}^{\infty}\dfrac{\pi n^3}{3^n}\), 且显然 \(\sum\limits_{n = 1}^{\infty}\dfrac{\pi n^3}{3^n}\) 收敛, 故由比较判别法知 \(\sum\limits_{n = 1}^{\infty}n^3\sin\dfrac{\pi}{3^n}\) 收敛.

第 3 题

判断下列级数的敛散性.

(2) \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{3^n}\left(1 + \dfrac{1}{n}\right)^{n^2}\);

(4) \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{\sqrt[3]{n+1}}\ln\dfrac{n + 2}{n}\);

(6) \(\sum\limits_{n = 1}^{\infty}\dfrac{\ln n!}{n!}\);

解

(2) 由于 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{3^n}\left(1 + \dfrac{1}{n}\right)^{n^2} \le \sum\limits_{n = 1}^{\infty}\left(\dfrac{e}{3}\right)^n\) 且显然 \(\sum\limits_{n = 1}^{\infty}\left(\dfrac{e}{3}\right)^n\) 收敛, 故由比较判别法知 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{3^n}\left(1 + \dfrac{1}{n}\right)^{n^2}\) 收敛.

(4) 由于 \(\lim\limits_{n \to \infty}n^{\frac{4}{3}}\cdot u_n = \lim\limits_{n \to \infty}\sqrt[3]{\dfrac{n}{n + 1}}\cdot n\ln\dfrac{n + 2}{n} = \lim\limits_{n \to \infty}\sqrt[3]{\dfrac{n}{n + 1}}\cdot \lim\limits_{n \to \infty}n\ln\dfrac{n + 2}{n} = 2\). 故由推论 5.2.1 知 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{\sqrt[3]{n+1}}\ln\dfrac{n + 2}{n}\) 收敛.

(6) 由于 \(\sum\limits_{n = 1}^{\infty}\dfrac{\ln n!}{n!} < \sum\limits_{n = 1}^{\infty}\dfrac{n}{n!} < 2 + \sum\limits_{n = 3}^{\infty}\dfrac{1}{(n - 1)(n - 2)}\), 且显然 \(\sum\limits_{n = 3}^{\infty}\dfrac{1}{(n - 1)(n - 2)}\) 收敛, 故由比较判别法知 \(\sum\limits_{n = 1}^{\infty}\dfrac{\ln n!}{n!}\) 收敛.

第 4 题

设 \(u_n, v_n > 0\), \(\dfrac{u_{n + 1}}{u_n} \le \dfrac{v_{n + 1}}{v_n}\), 证明: \(\sum\limits_{n = 1}^{\infty}v_n\) 收敛, 则 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛.

证明

由于 \(\dfrac{u_{n + 1}}{u_n} \le \dfrac{v_{n + 1}}{v_n}\), 故将左式从 \(n = 1\) 连续乘到 \(n = k\) 得到 \(\dfrac{u_{k + 1}}{u_1} \le \dfrac{v_{k + 1}}{v_1}\). 由于 \(\sum\limits_{n = 1}^{\infty}v_n\) 收敛, 故 \(\sum\limits_{n = 1}^{\infty}\dfrac{u_1}{v_1}v_n\) 收敛. 而 \(\sum\limits_{n = 1}^{\infty}u_{n} \le \sum\limits_{n = 1}^{\infty}\dfrac{u_1}{v_1}v_n\). 由比较判别法知 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛.

第 5 题

设 \(u_n > 0\), 数列 \(\{nu_n\}\) 有界, 证明: \(\sum\limits_{n = 1}^{\infty}\dfrac{u_n}{n}\) 收敛.

证明

设 \(nu_n \le M, \forall n \in \mathbb{N}^{*}\), 因此 \(u_n \le \dfrac{M}{n}\). 则 \(\sum\limits_{n = 1}^{\infty}\dfrac{u_n}{n} \le \sum\limits_{n = 1}^{\infty}\dfrac{M}{n^2}\). 显然 \(\sum\limits_{n = 1}^{\infty}\dfrac{M}{n^2}\) 收敛. 由比较判别法知 \(\sum\limits_{n = 1}^{\infty}\dfrac{u_n}{n}\) 收敛.

第 6 题

设 \(u_n, v_n > 0\), 且 \(\sum\limits_{n = 1}^{\infty}v_n, \sum\limits_{n = 1}^{\infty}u_n\) 均发散, 考察 \(\sum\limits_{n = 1}^{\infty}\max\{v_n, u_n\}\) 和 \(\sum\limits_{n = 1}^{\infty}\min\{v_n, u_n\}\) 的敛散性.

解

由于 \(\sum\limits_{n = 1}^{\infty}\max\{v_n, u_n\} \ge \sum\limits_{n = 1}^{\infty}u_n\) 且 \(\sum\limits_{n = 1}^{\infty}\max\{v_n, u_n\} \ge \sum\limits_{n = 1}^{\infty}v_n\), 故由比较判别法知 \(\sum\limits_{n = 1}^{\infty}\max\{v_n, u_n\}\) 发散.

取 \(u_n = v_n = n\), 则此时 \(\sum\limits_{n = 1}^{\infty}\min\{v_n, u_n\} = \sum\limits_{n = 1}^{\infty}n\) 发散. 取 \(u_n = \begin{cases}n, & n = 2k - 1, \\ \dfrac{1}{n^2}, & n = 2k\end{cases}, v_n = \begin{cases}\dfrac{1}{n^2}, & n = 2k - 1, \\ n, & n = 2k\end{cases}k \in \mathbb{N}^{*}\), 则此时 \(\sum\limits_{n = 1}^{\infty}\min\{v_n, u_n\} = \sum\limits_{n = 1}^{\infty}\dfrac{1}{n^2}\) 收敛. 因此 \(\sum\limits_{n = 1}^{\infty}\min\{v_n, u_n\}\) 既可能发散, 也可能收敛.

第 7 题

若 \(\sum\limits_{n = 1}^{\infty}u_n(u_n>0)\) 收敛, 则 \(\lim\limits_{n \to \infty}\dfrac{u_{n + 1}}{u_n} = l < 1\) 正确吗? 请举例说明.

解

不正确. 考虑 \(u_n = \dfrac{1}{n^2}\), 则 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛, 但 \(\lim\limits_{n \to \infty}\dfrac{u_{n + 1}}{u_n} = \lim\limits_{n \to \infty}\dfrac{\frac{1}{(n + 1)^2}}{\frac{1}{n^2}} = \lim\limits_{n \to \infty}\dfrac{n^2}{(n+1)^2} = 1\), 不小于 \(1\). 矛盾.

第 8 题

利用 Raabe 判别法, 讨论下列级数的敛散性.

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

(2) \(\sum\limits_{n = 1}^{\infty}\dfrac{n!n^{-p}}{q(q + 1)(q + 2)\cdots(q + n)}, p, q > 0\).

解

(1) \(n\left(\dfrac{u_n}{u_{n + 1}} - 1\right) = n\left(\dfrac{1 + \sqrt{n + 1}}{\sqrt{n + 1}} - 1\right) = \dfrac{n}{\sqrt{n + 1}}\). 取 \(\rho = 2\), 则当 \(n \ge 100\) 时有 \(\dfrac{n}{\sqrt{n + 1}} \ge \rho = 2\). 由 Raabe 判别法知 \(\sum\limits_{n = 1}^{\infty}\dfrac{\sqrt{n!}}{(1 + \sqrt{1})(1 + \sqrt{2})\cdots(1 + \sqrt{n})}\) 收敛.

(2)

\[ \begin{aligned} n\left(\dfrac{u_n}{u_{n + 1}} - 1\right) & = n\left(\dfrac{q + n + 1}{n + 1}\cdot \left(1 - \dfrac{1}{n + 1}\right)^{-p} - 1\right) \\ & = n\left(\dfrac{q + n + 1}{n + 1}\cdot\left(1 + \dfrac{p}{n + 1} + O\left(\dfrac{1}{(n + 1)^2}\right)\right) - 1\right) \\ & = \dfrac{n(p + q)}{n + 1} + O\left(\dfrac{1}{n}\right) \end{aligned} \]

因此, 由 Raabe 判别法知, 当 \(p + q > 1\) 时该级数收敛, 当 \(p + q \le 1\) 时该级数发散.

第 10 题

设 \(\sum\limits_{n = 1}^{\infty}u_n\) 为正项级数, 证明: \(\sum\limits_{n = 1}^{\infty}u_n\) 与 \(\sum\limits_{n = 1}^{\infty}\dfrac{u_n}{u_n + 1}\) 的敛散性相同.

证明

若 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛, 则 \(\lim\limits_{n \to \infty}u_n = 0\). 此时 \(\lim\limits_{n \to \infty} \dfrac{u_n}{\frac{u_n}{u_n + 1}} = \lim\limits_{n \to \infty}(u_n + 1) = 1\), 由比值判别法知 \(\sum\limits_{n = 1}^{\infty}\dfrac{u_n}{u_n + 1}\) 收敛.

若 \(\sum\limits_{n = 1}^{\infty}\dfrac{u_n}{u_n + 1}\) 收敛, 则 \(\lim\limits_{n \to \infty}\dfrac{u_n}{u_n + 1} = 0\), 这说明 \(\lim\limits_{n \to \infty}u_n = 0\). 此时 \(\lim\limits_{n \to \infty} \dfrac{u_n}{\frac{u_n}{u_n + 1}} = \lim\limits_{n \to \infty}(u_n + 1) = 1\), 由比值判别法知 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛.