习题5.1
第 2 题
级数 \(\sum\limits_{n = 1}^{\infty}u_n\) 的部分和序列 \(S_n\) 满足 \(\lim\limits_{n \to \infty}S_{2n + 1}\) 存在, \(\lim\limits_{n\to \infty}u_n = 0\), 证明: \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛.
证明
设 \(\lim\limits_{n\to\infty}S_{2n + 1} = L\). 由于 \(\lim\limits_{n\to\infty}S_{2n} = \lim\limits_{n \to \infty}(S_{2n + 1} - u_{2n + 1}) = \lim\limits_{n \to \infty}S_{2n + 1} - \lim\limits_{n \to \infty}u_{2n + 1} = L\), 故 \(\sum\limits_{n = 1}^{\infty}u_n = \lim\limits_{n\to\infty}S_{2n + 1} = \lim\limits_{n\to\infty}S_{2n} = L\), 级数收敛.
第 3 题
已知级数 \(\sum\limits_{n = 1}^{\infty}u_n\) 的部分和序列 \(S_n = \dfrac{2n}{n + 1}, n = 1, 2, \cdots\), 求:
(1) \(u_n\) 的通项公式;
(2) 判断 \(\sum\limits_{n = 1}^{\infty}u_n\) 的收敛性.
解
(1) 当 \(n = 1\) 时, \(u_1 = S_1 = 1\). 当 \(n \ge 2\) 时, \(u_n = S_{n} - S_{n - 1} = \dfrac{2n}{n + 1} - \dfrac{2(n - 1)}{n} = \dfrac{2}{n(n + 1)}\). 因此 \(u_n = \dfrac{2}{n(n + 1)}(n \in \mathbb{N^*})\).
(2) 由题知 \(\sum\limits_{n = 1}^{\infty}u_n = \lim\limits_{n \to \infty}S_n = \lim\limits_{n \to \infty}\dfrac{2n}{n + 1} = 2\). 该级数收敛.
第 4 题
已知级数 \(\sum\limits_{n = 1}^{\infty}u_n\) 中 \(u_n > 0\), 证明:
(1) \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛 \(\iff\) 级数 \((u_1 + \cdots + u_{n_1}) + (u_{n_1 + 1} + \cdots + u_{n_2}) + \cdots + (u_{n_{k-1} + 1} + \cdots + u_{n_k}) + \cdots\) 收敛;
(2) \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛 \(\implies\) \(\sum\limits_{n = 1}^{\infty}u_{2n + 1}\) 收敛.
证明
(1) \(\implies\): 设 \(\sum\limits_{n = 1}^{\infty}u_n\) 的部分和数列为 \(\{S_n\}\), 则新的级数的部分和数列为 \(\{S_{n_k}\}\), \(\{S_{n_k}\}\) 为 \(\{S_n\}\) 的子列, 有相同的收敛性, 且收敛到同一数.
\(\impliedby\): 取 \(n_k = k\), 则得证.
(2) 由于 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛, 所以 \(\forall \epsilon > 0, \exists N \in \mathbb{N^*}, \forall n > N, \forall p \in \mathbb{N^*}\), 有 \(\left|\sum\limits_{k = n + 1}^{n + p}u_k\right| < \epsilon\). 对 \(n'= 2n+2 > N, p' = 2p - 1 > 0\), 有 \(\left|\sum\limits_{k = 2n + 3}^{2n + 2p + 1}u_k\right| < \epsilon\). 因此 \(\left|\sum\limits_{k = n + 1}^{n + p}u_{2k + 1}\right| < \left|\sum\limits_{k = 2n + 3}^{2n + 2p + 1}u_k\right| < \epsilon\). 因此 \(\sum\limits_{n = 1}^{\infty}u_{2n + 1}\) 收敛.
第 5 题
设数列 \(\{u_n\}\) 满足 \(\lim\limits_{n \to \infty}nu_n = 0\), 证明: 级数 \(\sum\limits_{n = 1}^{\infty}(n+1)(u_{n+1}-u_n)\) 收敛等价于 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛.
证明
由于 \(\lim\limits_{n \to \infty}nu_n = 0\), 故 \(\exists N \in \mathbb{N}\), \(\forall \epsilon > 0, \forall n \ge N\), 有 \(\left|nu_n\right| < \dfrac{\epsilon}{3}\). \(\forall p > 0\), 显然 \(n + p \ge N\), 因此也有 \(\left|(n + p + 1)u_{n + p + 1}\right| < \dfrac{\epsilon}{3}\). 对上述 \(\epsilon > 0\), 考虑 \(\left|\sum\limits_{k = n + 1}^{n + p}(k + 1)(u_{k + 1} - u_k)\right| = \left|(n + p + 1)u_{n + p + 1} - (n + 1)u_{n + 1} - \sum\limits_{k = n + 1}^{n + p}u_k\right|\).
若 \(\sum\limits_{n = 1}^{\infty}(n+1)(u_{n+1}-u_n)\) 收敛, 则 \(\left|\sum\limits_{k = n + 1}^{n + p}(k + 1)(u_{k + 1} - u_k)\right| = \left|(n + p + 1)u_{n + p + 1} - (n + 1)u_{n + 1} - \sum\limits_{k = n + 1}^{n + p}u_k\right| < \epsilon\), 则 \(\left|\sum\limits_{k = n + 1}^{n + p}u_k\right| < 3\epsilon\). 这说明 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛.
反之, 若 \(\sum\limits_{n = 1}^{\infty}u_n\) 收敛, 则 \(\left|\sum\limits_{k = n + 1}^{n + p}(k + 1)(u_{k + 1} - u_k)\right| = \left|(n + p + 1)u_{n + p + 1} - (n + 1)u_{n + 1} - \sum\limits_{k = n + 1}^{n + p}u_k\right| < 3\epsilon\), 这说明 \(\sum\limits_{n = 1}^{\infty}(n+1)(u_{n+1}-u_n)\) 收敛.
命题得证.
第 6 题
利用定义判断下列级数的敛散性, 对收敛的级数求和.
(1) \(\sum\limits_{n = 1}^{\infty}100\left(\dfrac{1}{4}\right)^{n-1}\);
(3) \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{(2n-1)(2n + 3)}\);
(5) \(\sum\limits_{n = 1}^{\infty}\dfrac{(-1)^{n}n^3}{2n^3 + n}\);
(7) \(\sum\limits_{n = 1}^{\infty}\arctan{\dfrac{1}{2n^2}}\);
(9) \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{\sqrt[n]{n}}\).
解
(1) 由于 \(S_n = 100 \times 1 \times\dfrac{1 - \left(\frac{1}{4}\right)^{n}}{1 - \frac{1}{4}} = \dfrac{400}{3}\left(1 - 4^{-n}\right)\), 因此 \(\sum\limits_{n = 1}^{\infty}100\left(\dfrac{1}{4}\right)^{n-1} = \lim\limits_{n \to \infty}S_n = \dfrac{400}{3}\).
(3) 由于 \(S_n = \dfrac{1}{4}\sum\limits_{k = 1}^{n}\left(\dfrac{1}{2k - 1} - \dfrac{1}{2k + 3}\right) = \dfrac{1}{4}\left(\dfrac{4}{3} - \dfrac{1}{2n + 1} - \dfrac{1}{2n + 3}\right)\), 因此 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{(2n-1)(2n + 3)} = \lim\limits_{n\to\infty}S_n = \dfrac{1}{3}\).
(5) 由于 \(\lim\limits_{n \to \infty}u_n\) 不存在, 所以该级数发散.
(7) 由于 \(\arctan{\dfrac{1}{2n^2}} = \arctan{\dfrac{1}{2n - 1}} - \arctan{\dfrac{1}{2n + 1}}\), 所以 \(S_n = \sum\limits_{k = 1}^{n}\left(\arctan\dfrac{1}{2k - 1} - \arctan\dfrac{1}{2k + 1}\right) = \arctan 1 - \arctan\dfrac{1}{2n + 1} = \dfrac{\pi}{4} - \arctan\dfrac{1}{2n + 1}\). 故 \(\sum\limits_{n = 1}^{\infty}\arctan{\dfrac{1}{2n^2}} = \lim\limits_{n \to \infty}S_n = \dfrac{\pi}{4}\).
(9) 由于 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{\sqrt[n]{n}} \ge \sum\limits_{n = 1}^{\infty}\dfrac{1}{n}\), 且后者发散, 故级数发散.
第 7 题
求级数 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{n(n+m)}(m > 0, m \in \mathbb{Z})\) 的和.
解