# Quod Erat Demonstrandum

## 2014/05/16

### 證明某級數

Filed under: Fun,Pure Mathematics — johnmayhk @ 11:00 下午
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$\frac{1}{x^3(x+1)^3}$

$\frac{1}{x^3(x+1)^3}\equiv 6(\frac{1}{x}-\frac{1}{x+1})-3(\frac{1}{x^2}+\frac{1}{(x+1)^2})+(\frac{1}{x^3}-\frac{1}{(x+1)^3})$

$\displaystyle \sum_{k=1}^n\frac{1}{k^3(k+1)^3}$

$=\displaystyle \sum_{k=1}^n(6(\frac{1}{k}-\frac{1}{k+1})-3(\frac{1}{k^2}+\frac{1}{(k+1)^2})+(\frac{1}{k^3}-\frac{1}{(k+1)^3}))$

$=6(1-\frac{1}{n+1})-3\displaystyle \sum_{k=1}^n(\frac{1}{k^2}+\frac{1}{(k+1)^2})+(1-\frac{1}{(n+1)^3})$

$\displaystyle \lim_{n\rightarrow \infty}\sum_{k=1}^n\frac{1}{k^3(k+1)^3}$

$=\displaystyle \lim_{n\rightarrow \infty}(6(1-\frac{1}{n+1})-3\displaystyle \sum_{k=1}^n(\frac{1}{k^2}+\frac{1}{(k+1)^2})+(1-\frac{1}{(n+1)^3}))$

$=6-3(\frac{\pi^2}{6}+\frac{\pi^2}{6}-1)+1$

(因為 $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$)

$\frac{1}{1^3\cdot 2^3}+\frac{1}{2^3\cdot 3^3}+\frac{1}{3^3\cdot 4^3}+\dots =10-\pi^2$

$\frac{1}{x^3(x+1)^3}\equiv \frac{a_1}{x}+\frac{a_2}{x^2}+\frac{a_3}{x^3}+\frac{b_1}{x+1}+\frac{b_2}{(x+1)^2}+\frac{b_3}{(x+1)^3}$……………. (*)

$\frac{1}{x^3(x+1)^3}\equiv \frac{a_1x^2+a_2x+a_3}{x^3}+g(x)$

（其中 $g(x)=\frac{b_1}{x+1}+\frac{b_2}{(x+1)^2}+\frac{b_3}{(x+1)^3}$

$\frac{1}{(x+1)^3}\equiv a_1x^2+a_2x+a_3+x^3g(x)$

$\frac{-3}{(x+1)^4}\equiv 2a_1x+a_2+x^3g'(x)+3x^2g(x)$

$\frac{12}{(x+1)^5}\equiv 2a_1+x^3g''(x)+3x^2g'(x)+3x^2g'(x)+6xg(x)$

$b_3=-1,b_2=-3,b_1=-6$

$\frac{1}{(x-a)^n(x-b)^n}\equiv \frac{(-1)^n}{(a-b)^{2n}}\displaystyle \sum_{k=0}^{n-1}\frac{(n+k-1)!}{k!(n-1)!}((\frac{b-a}{x-a})^{n-k}+(\frac{a-b}{x-b})^{n-k})$