$∑_{k=1}^n1/\sqrt k∼2\sqrt n$

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Mathematical Analysis, Apostol著, 211页, Exercise 8.8
Let $a_n=2\sqrt n-∑_{k=1}^n1/\sqrt k$. Prove that the sequence $\{a_n\}$ converges to a limit $p$ in the interval $1<p<2$.
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Solution 1.
$2\sqrt{k}-2\sqrt{k-1} = \frac{2}{\sqrt{k}+\sqrt{k-1}}$
$⇒1/\sqrt{k-1}\overset{\text{(1)}}>2\sqrt{k}-2\sqrt{k-1}\overset{\text{(2)}}>1/\sqrt{k}$
Summing (1) over $k=2,⋯,n+1$,
$∑_{k=2}^{n+1}1/\sqrt{k-1}>2\sqrt n-2⇒2\sqrt n-∑_{k=1}^n1/\sqrt k<2$
Summing (2) over $k=2,⋯,n$,
$∑_{k=2}^n1/\sqrt k<2\sqrt n-2⇒2\sqrt n-∑_{k=1}^n1/\sqrt k>1$
From (2), the sequence $(a_n)$ is increasing, but we proved $(a_n)$ is bounded above by 2, so $(a_n)$ converges.

Solution 2.
Use $\int_1^{n+1}x^{-1/2}\,\mathrm dx$ to derive the inequality. The rest is same.

战巡

抄一段《特殊函数概论》里的做法好了

由欧拉-麦克劳林公式，会有任意函数$F(x)$，有
\[\int_a^{a+mh}F(x)dx=h\left[\frac{F(a)}{2}+F(a+h)+...+F(a+(m-1)h)+\frac{F(a+mh)}{2}\right]\]
\[+\sum_{k=1}^n\frac{(-1)^kB_k h^{2k}}{(2k)!}[F^{(2k-1)}(a+mh)-F^{(2k-1)}(a)]+R_n\]
其中
\[R_n=h^{2n+1}\int_0^mP_{2n}(t)F^{(2n)}(a+ht)dt\]
这里$P_{\lambda}(t)=\frac{\varphi_{\lambda}(t)}{\lambda!}$，$\varphi_{\lambda}(t)$为伯努利多项式

此时令$h=1,m\to\infty,F(x)=x^{-s}$，会得到
\[\int_a^\infty\frac{1}{x^s}dx=\frac{1}{(s-1)a^{s-1}}=\frac{1}{2}\cdot\frac{1}{a^s}+\frac{1}{(a+1)^s}+\frac{1}{(a+2)^s}+...\]
\[+\sum_{k=1}^n\frac{(-1)^{k-1}B_k\frac{(s+2k-2)!}{(s-1)!}}{(2k)!a^{s+2k-1}}+R_n\]
那么
\[\zeta(s,a+1)=\frac{1}{(a+1)^s}+\frac{1}{(a+2)^s}+...=\frac{1}{s-1}\cdot\frac{1}{a^{s-1}}-\frac{1}{2a^s}+\sum_{k=1}^n\frac{(-1)^{k-1}B_k\frac{(s+2k-2)!}{(s-1)!}}{(2k)!a^{s+2k-1}}+R_n\]

如此就有
\[\frac{1}{1-s}\cdot\frac{1}{n^{s-1}}-\sum_{k=1}^n\frac{1}{k^s}=\frac{1}{1-s}\cdot\frac{1}{n^{s-1}}-\left[\sum_{k=1}^\infty\frac{1}{k^s}-\sum_{k=n+1}^\infty\frac{1}{k^s}\right]\]
\[=\frac{1}{1-s}\cdot\frac{1}{n^{s-1}}-[\zeta(s)-\zeta(s,n+1)]\]
\[=-\zeta(s)-\frac{1}{2n^s}+\sum_{k=1}^n\frac{(-1)^{k-1}B_k\frac{(s+2k-2)!}{(s-1)!}}{(2k)!n^{s+2k-1}}+R_n\]
在$n\to\infty$时后面全是$0$，也就是有
\[\lim_{n\to\infty}\left[\frac{1}{1-s}\cdot\frac{1}{n^{s-1}}-\sum_{k=1}^n\frac{1}{k^s}\right]=-\zeta(s)\]

套进去$s=\frac{1}{2}$，就有
\[\lim_{n\to\infty}\left[2\sqrt{n}-\sum_{k=1}^n\frac{1}{\sqrt{k}}\right]=-\zeta(\frac{1}{2})=1.46035...\]

上面同时还能得到一个蛮有意思的结论：
\[\lim_{n\to\infty}\left[\frac{n}{s-1}-\sum_{k=n+1}^\infty\left(\frac{n}{k}\right)^s\right]=\frac{1}{2}\]