อ้างอิง:
ข้อความเดิมเขียนโดยคุณ passer-by
30.Find $$ \sum_{n=1}^{\infty} (-1)^n ( \ln 2 - \frac{1}{n+1} - \frac{1}{n+2}- \cdots -\frac{1}{2n}) $$
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$$S=\sum_{n=1}^{\infty}(-1)^n\left(\ln 2-\frac{1}{n+1}-\frac{1}{n+2}-\cdots -\frac{1}{2n}\right)=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}(-1)^k\left(\ln 2-\left(H(2k)-H(k)\right)\right)$$
Let $a_{n}=\ln 2-\left(H(2n)-H(n)\right)$,Now observe that
$$\frac{1}{n+1}<\frac{1}{2n+1}+\frac{1}{2n+2}$$
$$H(2n)-H(n)<H(2n)+\frac{1}{2n+1}+\frac{1}{2n+2}-H(n)-\frac{1}{n+1}$$
$$a_{n}=\ln 2-\left(H(2n)-H(n)\right)>\ln 2-\left(H(2n+2)-H(n+1)\right)=a_{n+1},\lim_{n\rightarrow\infty}a_{n}=0$$
$\displaystyle{\therefore\sum_{k=1}^{\infty}(-1)^{k}a_{k}=S}$ converges by alternating series test.
We can easily see that
$$\lim_{n\rightarrow\infty}\left[\sum_{k=1}^{2n+1}(-1)^{k}a_{n}\right]=\lim_{n\rightarrow\infty}\left[-a_{2n+1}+\sum_{k=1}^{2n}(-1)^{k}a_{n}\right]=\lim_{n\rightarrow\infty}\sum_{k=1}^{2n}(-1)^{k}a_{n}=S$$
To finish the problem we can do it this way...
$$S=\lim_{n\rightarrow\infty}\sum_{k=1}^{2n}(-1)^{k}a_{n}=\lim_{n\rightarrow\infty}\sum_{k=1}^{2n}(-1)^{k}\left[\ln 2-H(2k)+H(k)\right]=\lim_{n\rightarrow\infty}\left[\sum_{k=1}^{2n}(-1)^{k}H(k)-\sum_{k=1}^{2n}(-1)^{k}H(2k)\right]$$
$$S=\lim_{n\rightarrow\infty}\left[\left(\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2n}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{7}+\frac{1}{8}+\cdots +\frac{1}{4n-1}+\frac{1}{4n}\right)\right]$$
$$S=-\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left[\int_{0}^{1}x^{4n-2}-x^{4n-1}\right]=-\int_{0}^{1}\frac{x^{2}-x^{3}}{1-x^{4}}dx=\frac{\pi-6\ln 2}{8}$$
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