今天想的一題高微

在 2004/02/20 00:49:48 於未來最舊小棧留下的問題，終於解了。

題目

Prove

$$a_n = 2\sqrt{n} - \sum^n_{k=1}\frac{1}{\sqrt{k}}\to p, 1<p<2$$

解答

$\lbrace a_n\rbrace$ is monotone increasing, because

$$\begin{align*} a_{n+1} - a_{n} &= 2\sqrt{n+1} - 2\sqrt{n} - \frac{1}{\sqrt{n+1}}\\ &=\frac{2}{\sqrt{n+1}+\sqrt{n}}- \frac{1}{\sqrt{n+1}}\\ &>\frac{2}{\sqrt{n+1}+\sqrt{n+1}}- \frac{1}{\sqrt{n+1}}=0 \end{align*}$$

Since $\frac{1}{\sqrt{x}}$ is a Convex function and by Trapezoidal rule, we have

$$\sum^n_{k=1}\frac{1}{\sqrt{k}}-\frac{1}{2}+\frac{1}{\sqrt{n+1}}=\frac{1}{2}\sum^n_{k=1}(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}})\geq\int^{n+1}_{1}\frac{1}{\sqrt{k}}\mathrm{d}k=2\sqrt{n+1}-2$$

then

$$\begin{aligned} &2\sqrt{n}-2\sqrt{n+1}+2-\frac{1}{2}+\frac{1}{\sqrt{n+1}}\geq2\sqrt{n}-\sum^n_{k=1}\frac{1}{\sqrt {k}}\\ &\implies\frac{3}{2}-(2\sqrt{n+1} - 2\sqrt{n} - \frac{1}{\sqrt{n+1}})\geq a_n\\ &\implies\frac{3}{2}\geq a_n+a_{n+1}-a_n=a_{n+1}>a_1=1 \end{aligned}$$

Hence by Monotone convergence theorem, we have

$$1 < \lim_{n\to\infty}a_n\leq\frac{3}{2}<2$$

後記

當年的做法跟 這篇 發文者一樣，只走到 $1 < \lim\limits_{n\to\infty}a_n\leq2$，後來就不了了之。