Quod Erat Demonstrandum

2015/08/28

sum of 1/k^2 from 1 to infinity

Filed under: Additional / Applied Mathematics,Fun — johnmayhk @ 4:30 下午
Tags:

非正式地所謂證明 \displaystyle \sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}
…………………………………………….

首先要知,下式的根之總和(sum of roots)

ax^n+bx^{n-1}+\dots +cx+d=0 ………. (*)

-\frac{b}{a}

把 (*) 除以 x^n,得

a+b(\frac{1}{x})+\dots +c(\frac{1}{x})^{n-1}+d(\frac{1}{x})^n=0

y=\frac{1}{x},得

dy^n+cy^{n-1}+\dots +by+a=0

並知上式的根,就是 (*) 的根之倒數(reciprocal)。

所以 (*) 的根之倒數的總和(sum of reciprocal of roots)

就是 -\frac{c}{d}

…………………………………………….

好了,「偽證明」開始。

因為

\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots

所以

\sin(\sqrt{x})=\sqrt{x}-\frac{x\sqrt{x}}{3!}+\frac{x^2\sqrt{x}}{5!}-\dots

\Rightarrow \frac{\sin(\sqrt{x})}{\sqrt{x}}=1-\frac{x}{3!}+\frac{x^2}{5!}-...

現在是「夾硬來」:把上式的右邊視為「無限大 degree 的多項式」。

故上式右邊的根之倒數總和(sum of reciprocal of roots)是

-\frac{-1/3!}{1}=\frac{1}{6}

而上式左邊的根是 k^2\pi^2(其中 k 是正整數),故上式左邊的根之倒數總和是

\displaystyle \sum_{k=1}^\infty \frac{1}{k^2\pi^2}

於是有

\displaystyle \sum_{k=1}^\infty \frac{1}{k^2\pi^2}=\frac{1}{6}

\displaystyle \sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}

johnmayhk-thisisnotscience

2 則迴響 »

  1. Who wrote this proof? Where is wrong? Sorry for my stupidity.

    迴響 由 Simon YAU YAU — 2015/08/30 @ 6:00 下午 | 回覆


RSS feed for comments on this post. TrackBack URI

發表迴響

在下方填入你的資料或按右方圖示以社群網站登入:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / 變更 )

Twitter picture

You are commenting using your Twitter account. Log Out / 變更 )

Facebook照片

You are commenting using your Facebook account. Log Out / 變更 )

Google+ photo

You are commenting using your Google+ account. Log Out / 變更 )

連結到 %s

在WordPress.com寫網誌.

%d 位部落客按了讚: