###### tags: `數學問題`
# Kevin問題
$\begin{align}\end{align}$
1. $\begin{align}\sum\limits_{n=2}^\infty(\frac{1}{(n^2+n)^2}-\frac{2}{(n+1)^2})\end{align}$
注意到 $n=2$，意思是可以容許有 $(n-1)$
$\begin{align}\frac{1}{(n^2+n)^2}-\frac{2}{(n+1)^2}=\frac{1-2n^2}{n^2(n+1)^2}=\frac{(n^4-2n^2+1)-n^4}{n^2(n+1)^2}\end{align}$
$\begin{align}=\frac{(n^2-1)^2-n^4}{n^2(n+1)^2}=\frac{(n-1)^2}{n^2}-\frac{n^2}{(n+1)^2}\end{align}$
原式 $\begin{align}=\lim\limits_{k\to\infty}\sum\limits_{n=2}^k(\frac{(n-1)^2}{n^2}-\frac{n^2}{(n+1)^2})\end{align}$
$\begin{align}=\lim\limits_{k\to\infty}\left[(\frac{1^2}{2^2}+\frac{2^2}{3^2}+\cdots+\frac{(k-1)^2}{k^2})-(\frac{2^2}{3^2}+\cdots\frac{(k-1)^2}{k^2}+\frac{k^2}{(k+1)^2})\right]\end{align}$
$\begin{align}=\lim\limits_{k\to\infty}(\frac{1^2}{2^2}-\frac{k^2}{(k+1)^2})=\frac{1}{4}-1=-\frac{3}{4}\end{align}$
2. 求 $\begin{align}\sum\limits_{n=1}^\infty\frac{1}{2^{n+(-1)^\frac{n(n+1)}{2}}}\end{align}$
首先，判斷 $\begin{align}\frac{n(n+1)}{2}\end{align}$ 的奇偶性，
當 $n=4k、4k+3$ 時為偶；當 $n=4k+1、4k+2$ 時為奇
所求即為 $\begin{align}\frac{1}{2^{1-1}}+\frac{1}{2^{2-1}}+\frac{1}{2^{3+1}}+\frac{1}{2^{4+1}}+\frac{1}{2^{5-1}}+\frac{1}{2^{6-1}}+\frac{1}{2^{7+1}}+\frac{1}{2^{8+1}}+\cdots\end{align}$
$\begin{align}=\frac{1+\frac{1}{2}+\frac{1}{16}+\frac{1}{32}}{1-\frac{1}{2^4}}=\frac{32+16+2+1}{32-2}=\frac{51}{30}=\frac{17}{10}\end{align}$

3. 已知 $n\in\mathbb{N}$，當 $\begin{align}\frac{1}{2^n}<x\leq\frac{1}{2^{n-1}}\end{align}$ 時，函數 $\begin{align}f(x)=a_n(\log_{\frac{1}{2}}x)^n\end{align}$ 且 $a_1=1$，若 $f(x)$ 為區間 $(0,1]$ 上連續函數，求 $a_n$
$\begin{align}\lim\limits_{x\to\frac{1}{2^n}^+}f(x)=\lim\limits_{x\to\frac{1}{2^n}^+}a_n(\log_\frac{1}{2}x)^n=a_n(\log_\frac{1}{2}(\frac{1}{2^n}))^n=a_n\cdot n^n\end{align}$
$\begin{align}\lim\limits_{x\to\frac{1}{2^n}^-}f(x)=\lim\limits_{x\to\frac{1}{2^n}^-}a_{n+1}(\log_\frac{1}{2}x)^{n+1}=a_{n+1}(\log_\frac{1}{2}(\frac{1}{2^n}))^{n+1}=a_{n+1}\cdot n^{n+1}\end{align}$
因為 $f(x)$ 連續，所以 $a_n\cdot n^n=a_{n+1}\cdot n^{n+1}$
$\begin{align}a_n=na_{n+1}\end{align}$
由遞迴數列的累乘法，$a_n=\frac{1}{(n-1)!}$