西南财经大学 2020年数学分析第0题

注意到分母 $n^2-1 = (n-1)(n+1)$，设 $\frac{1}{(n-1)(n+1)} = \frac{A}{n-1} + \frac{B}{n+1}$，解得 $A = \frac12, B = -\frac12$，因此 $$ \frac{1}{n^2-1} = \frac12\left(\frac{1}{n-1} - \frac{1}{n+1}\right). $$ 原级数化为 $$ \sum_{n=2}^\infty (-1)^n \frac{1}{2^n} \cdot \frac12 \left( \frac{1}{n-1} - \frac{1}{n+1} \right) = \frac12 \sum_{n=2}^\infty (-1)^n \frac{1}{2^n} \left( \frac{1}{n-1} - \frac{1}{n+1} \right). $$

令 $S_1 = \sum_{n=2}^\infty (-1)^n \frac{1}{2^n} \cdot \frac{1}{n-1}$，$S_2 = \sum_{n=2}^\infty (-1)^n \frac{1}{2^n} \cdot \frac{1}{n+1}$，则原级数 $= \frac12(S_1 - S_2)$。 对 $S_1$：令 $k = n-1$，则 $n = k+1$，$k$ 从 $1$ 开始， $$ S_1 = \sum_{k=1}^\infty (-1)^{k+1} \frac{1}{2^{k+1}} \cdot \frac{1}{k} = -\frac12 \sum_{k=1}^\infty \frac{(-1)^k}{k} \left(\frac12\right)^k. $$ 对 $S_2$：令 $m = n+1$，则 $n = m-1$，$m$ 从 $3$ 开始， $$ S_2 = \sum_{m=3}^\infty (-1)^{m-1} \frac{1}{2^{m-1}} \cdot \frac{1}{m} = -2 \sum_{m=3}^\infty \frac{(-1)^m}{m} \left(\frac12\right)^m. $$

已知对数级数展开：对 $|x|<1$，$\sum_{n=1}^\infty \frac{(-1)^n x^n}{n} = -\ln(1+x)$。 取 $x = \frac12$，得 $$ \sum_{k=1}^\infty \frac{(-1)^k (1/2)^k}{k} = -\ln\left(1+\frac12\right) = -\ln\frac32. $$ 因此 $$ S_1 = -\frac12 \cdot \left( -\ln\frac32 \right) = \frac12 \ln\frac32. $$

先求全体和：$\sum_{m=1}^\infty \frac{(-1)^m (1/2)^m}{m} = -\ln\frac32$。 前两项：$m=1$ 项为 $-\frac12$，$m=2$ 项为 $\frac18$，故前两项和为 $-\frac12 + \frac18 = -\frac38$。 因此从 $m=3$ 开始的和为 $$ \sum_{m=3}^\infty \frac{(-1)^m (1/2)^m}{m} = \left(-\ln\frac32\right) - \left(-\frac38\right) = -\ln\frac32 + \frac38. $$ 代入 $S_2$ 表达式： $$ S_2 = -2 \left( -\ln\frac32 + \frac38 \right) = 2\ln\frac32 - \frac34. $$

原级数 $= \frac12 (S_1 - S_2)$，代入得 $$ \begin{aligned} \text{原级数} &= \frac12\left[ \frac12 \ln\frac32 - \left(2\ln\frac32 - \frac34\right) \right] \\ &= \frac12\left( \frac12\ln\frac32 - 2\ln\frac32 + \frac34 \right) \\ &= \frac12\left( -\frac32 \ln\frac32 + \frac34 \right) \\ &= -\frac34 \ln\frac32 + \frac38. \end{aligned} $$ 因此级数和为 $\displaystyle \frac{3}{8} - \frac{3}{4}\ln\frac{3}{2}$。