Proof that sum^infty of 1/n^2 = ...

**ragnar** · August 7th 2010, 11:34 AM

I am trying to prove $\frac{x^{2}}{2} = \pi x - \frac{\pi^{2}}{3} + 2\displaystyle \sum^{\infty}_{n = 1} \frac{cos(nx)}{n^{2}}$ for $0 \leq x \leq 2 \pi$ . I've done quite a lot:

Taking derivative of both sides and using part (a), we see that $\frac{x^{2}}{2}$ and $\pi x + 2\displaystyle \sum^{\infty}_{n = 1} \frac{cos(nx)}{n^{2}}$ differ by at most a constant on $0 < x < 2 \pi$ (in part (a) I have basically shown that the derivative of each side is equal). Evaluating at $\pi$ , we see that the constant difference is $\frac{\pi^{2}}{3}$ , which proves the equality. Likewise evaluating at 0 and $2 \pi$ we obtain the identity.

(Note, this argument depends on the claim that $\displaystyle \sum^{\infty}_{n = 1} \frac{1}{n^{2}} = \frac{\pi}{6}$ , at the point when I evaluate at $\pi$ . ...

So here's where I'm stuck. I have a hint that I'm supposed to use the following theorem:

If a function $f$ is continuous and periodic on $[0, 2 \pi]$ with period $2 \pi$ , then, using the usual notation about $a_{n}$ and $b_{n}$ , $\frac{1}{\pi} \displaystyle \int^{2 \pi}_{0} |f(t)|^{2}dt = \frac{a_{0}^{2}}{2} + \sum^{\infty}_{n = 1}(a_{n}^{2} + b_{n}^{2})$ .

My first question is, since my function isn't periodic, how do I use this theorem?

Anyway, I started trying this for the function I'm working with, but what I get is: $\frac{1}{\pi} \displaystyle \int^{2 \pi}_{0}\frac{x^{4}}{4}dx =$<br /> <br /> $\displaystyle \Bigl( \frac{1}{\pi} \int^{2 \pi}_{0} \frac{x^{2}}{2}dx \Bigr)^{2}/2 + \sum^{\infty}_{n = 1} \Bigl( \frac{1}{\pi} \int^{2 \pi}_{0} \frac{x^{2}}{2}cos(nx) dx \Bigr)^{2} + \Bigl( \frac{1}{\pi} \int^{2 \pi}_{0} \frac{x^{2}}{2}sin(nx) dx \Bigr)^{2}$

Now I have to say, already this looks messed up. Am I doing something wrong?

**Ackbeet** · August 7th 2010, 11:55 AM

My first question is, since my function isn't periodic, how do I use this theorem?

The answer is that your function is periodic. Or you can make it periodic. Just think of copies of $x^{2}/2$ repeating their values every $2\pi.$ Remember that your original function's domain was restricted to $[0,2\pi]$ . Therefore, it's not the same function as $x^{2}/2$ for all real numbers $x$ .

That's at least a partial answer, I think. Can you continue?

**ragnar** · August 7th 2010, 12:14 PM

I had a hunch that was why the theorem would hold on the interval, but I'm still worried about the rest. If I crunch some numbers on a black-board to compute what I have at the end of my last message, I get $\frac{4\pi^{3}}{5} = 2\pi + \frac{1}{\pi^{2}} \displaystyle \sum^{\infty}_{n = 1} \Bigl( \int^{2 \pi}_{0} \frac{x^{2}}{2}(cos^{2}(nx) + sin^{2}(nx)) \Bigr)^{2}$ , and the sum seems to diverge... And here, I don't even know for a fact that the function I should be putting into this theorem is the one I've been using, since after all, my goal is just to compute $\displaystyle \sum^{\infty}_{n = 1} \frac{1}{n^{2}}$ ...

**chisigma** · August 7th 2010, 12:24 PM

Why don't develop in Fourier series the function...

$\displaystyle \frac{\pi^{2}}{6} - \frac{\pi}{2}\ x + \frac{x^{2}}{4} , 0 < x < 2 \pi$ (1)

Kind regards

$\chi$ $\sigma$

**chiph588@** · August 7th 2010, 01:24 PM

Originally Posted by ragnar

I had a hunch that was why the theorem would hold on the interval, but I'm still worried about the rest. If I crunch some numbers on a black-board to compute what I have at the end of my last message, I get $\frac{4\pi^{3}}{5} = 2\pi + \frac{1}{\pi^{2}} \displaystyle \sum^{\infty}_{n = 1} \Bigl( \int^{2 \pi}_{0} \frac{x^{2}}{2}(cos^{2}(nx) + sin^{2}(nx)) \Bigr)^{2}$ , and the sum seems to diverge... And here, I don't even know for a fact that the function I should be putting into this theorem is the one I've been using, since after all, my goal is just to compute $\displaystyle \sum^{\infty}_{n = 1} \frac{1}{n^{2}}$ ...

First observe $\displaystyle \int_0^1\int_0^1(xy)^{n-1}dxdy = \frac1{n^2}$ .

Therefore $\displaystyle \zeta(2)=\sum_{n=1}^\infty \frac1{n^2}=\sum_{n=1}^\infty \int_0^1\int_0^1(xy)^{n-1}dxdy = \int_0^1\int_0^1\left( \sum_{n=1}^\infty (xy)^{n-1} \right)dxdy=\int_0^1\int_0^1\frac1{1-xy}dxdy$ , as $\displaystyle xy<1$ .
We can switch the order here as this converges absolutely.

Perform the substitution $\displaystyle (u,v)=\left( \frac{x+y}{2},\frac{y-x}{2} \right) \iff (x,y)=(u-v,\,u+v)$ .

We then obtain $\displaystyle \zeta(2)=2\iint_{D}\frac1{1-u^2+v^2}dvdu$ , where $\displaystyle D$ is the diamond with corners at $\displaystyle (0,0),\;\left( \frac12,\frac12 \right),\;\left( \frac12,-\frac12 \right),\;(1,0)$ .

Evaluating the integral we get $\displaystyle \zeta(2)=4\left( \int_0^{\frac12}\int_0^u\frac1{1-u^2+v^2}dvdu+\int_{\frac12}^{1}\int_0^{1-u}\frac1{1-u^2+v^2}dvdu \right)$ .

Skipping the mess of integration (it's straightforward, and Wolfram|Alpha can help if need be), $\displaystyle \zeta(2)=4\left( \frac{\pi^2}{72}+\frac{\pi^2}{36} \right)=\frac{\pi^2}6$ .

**ragnar** · August 7th 2010, 01:29 PM

I take it that's a hint that I'm approaching this the wrong way... [Edit: Nevermind, looks like a much fuller post now.]

**tonio** · August 7th 2010, 02:11 PM

Originally Posted by ragnar

I take it that's a hint that I'm approaching this the wrong way... [Edit: Nevermind, looks like a much fuller post now.]

Let's be classic: develop periodically $f(x)=x^2\,\,\,in\,\,\,[-\pi,\pi]$ . Note that this function and its

derivative are continuous everywhere and thus its values are given by its Fourier series. Moreover, since we've

an even funcion only cosines will appear in its F. series:

$a_0=\frac{1}{\pi}\int\limits_{-\pi}^\pi x^2dx=\frac{2\pi^2}{3}$

By parts twice, we get for $n\geq 1\,,\,\,a_n=\frac{1}{\pi}\int\limits_{-\pi}^\pi x^2\cos nxdx=-\frac{2}{\pi n}\int\limits_{-\pi}^\pi x\sin xdx=\frac{4\cos n\pi}{n^2}$ , so:

$x^2=\frac{a_0}{2}+\sum\limits^\infty_{n=1}a_n\cos nx=\frac{\pi^2}{3}+\sum\limits^\infty_{n=1}\frac{4 \cos n\pi \cos nx}{n^2}=\frac{\pi^2}{3}+4\sum\limits^\infty_{n=1} \frac{\cos n\pi\cos nx}{n^2}$ , and now choose $x=\pi$ and voilá!

Tonio

**ragnar** · August 7th 2010, 06:21 PM

I'm afraid I still don't see it. When I plug in pi, I get $\pi^{2} = \frac{\pi^{2}}{3} +\displaystyle 2\sum^{\infty}_{n=1} \frac{1-cos(2n\pi)}{n^{2}}$ , what is in the sum is then 0 and I get a contradiction.

... Oops, wrong trig law. Still computing...

**simplependulum** · August 7th 2010, 06:35 PM

I've just thought of a method but it is still advanced , actually i am looking for an elementary and geometric approach

.

We know , by expanding power series ,

$\int_1^{\infty} \frac{\ln(x)}{ x^2 - 1 }~dx = \frac{3}{4} \zeta(2)$

Then we are going to show that $\int_1^{\infty} \frac{\ln(x)}{ x^2 - 1 }~dx = \frac{\pi^2}{8}$ .

First thing , perform Euler's substitution , but the inverse substitution :

$x = t + \sqrt{t^2 + 1 }$

$dx = \frac{t + \sqrt{t^2+1}}{\sqrt{t^2 + 1 }} ~dt$

The integral becomes :

$\int_0^{\infty} \frac{ \ln( t + \sqrt{t^2 + 1 } ) }{ (2t^2 + 2t\sqrt{t^2+1} + 1 )- 1 } ~ \cdot ~ \frac{t + \sqrt{t^2+1}}{\sqrt{t^2 + 1 }} ~dt$

$= \int_0^{\infty} \frac{ \ln( t + \sqrt{t^2 + 1 } ) }{ 2t \sqrt{t^2 + 1 } } ~dt$

$= \frac{1}{4} \int_0^{\infty} \frac{ \ln[ (t + \sqrt{t^2 + 1 } )^2 ] }{ t \sqrt{t^2 + 1 } } ~dt$

$= \frac{1}{4} \int_0^{\infty} \ln{ \left[ \frac{ \sqrt{t^2 + 1 } + t }{ \sqrt{t^2 + 1 } - t } \right] }~\cdot ~ \frac{ dt }{ t \sqrt{t^2 + 1 } }$

$= \frac{1}{2} \int_0^{\infty} \int_0^1 \frac{dydt}{ 1+t^2 - t^2y^2 }$

$= \frac{1}{2} \int_0^1 \left[ \tan^{-1}(t\sqrt{1-y^2}) \right]_0^{\infty} ~\frac{dy}{\sqrt{1-y^2} }$

$= \frac{\pi}{4} \int_0^1 \frac{dy}{\sqrt{1-y^2 } }$

$= \frac{\pi^2}{8}$

**chiph588@** · August 7th 2010, 10:13 PM

Originally Posted by ragnar

I'm afraid I still don't see it. When I plug in pi, I get $\displaystyle \pi^{2} = \frac{\pi^{2}}{3} + 2\sum^{\infty}_{n=1} \frac{1-\cos(2n\pi)}{n^{2}}$ , what is in the sum is then 0 and I get a contradiction.

... Oops, wrong trig law. Still computing...

The numerator for your summand should be $\displaystyle \cos(n\pi)^2=1$ .

**mr fantastic** · August 7th 2010, 10:37 PM

Originally Posted by ragnar

[snip]
my goal is just to compute $\displaystyle \sum^{\infty}_{n = 1} \frac{1}{n^{2}}$ ...

Read this: http://scipp.ucsc.edu/~haber/archive...06/Sixways.pdf

**ragnar** · August 8th 2010, 09:46 AM

When I put in $\frac{1}{\pi}\int\limits_{-\pi}^\pi x^2\cos nxdx$ into Wolfram I get something very different from what you got, Tonio.

I looked at the link, Mr. Fantastic, but the part which gives the value of the sum just sort of states the result and I don't get the argument.

And I appreciate the many responses, but a lot of them use techniques that seem just too advanced, or they expand a Fourier series and even if I can blindly crunch the numbers I still have no idea why I should have thought of that particular series and so I feel as though I still don't understand what's going on.

**tonio** · August 8th 2010, 09:51 AM

Originally Posted by ragnar

When I put in $\frac{1}{\pi}\int\limits_{-\pi}^\pi x^2\cos nxdx$ into Wolfram I get something very different from what you got, Tonio.

I really don't care since I don't know what algorithms/programs is Wolfram implementing. The interesting question is what do you get when you do by parts (twice, as said) that integral?!
Note also that you can multiply by two the integral and integrate from zero to pi (why??)

Tonio

I looked at the link, Mr. Fantastic, but the part which gives the value of the sum just sort of states the result and I don't get the argument.

And I appreciate the many responses, but a lot of them use techniques that seem just too advanced, or they expand a Fourier series and even if I can blindly crunch the numbers I still have no idea why I should have thought of that particular series and so I feel as though I still don't understand what's going on.

.

**ragnar** · August 8th 2010, 10:16 AM

Ah, I thought I got something different from you too, but when I computed the last steps it came out. Okay, this works, thank you.

**mr fantastic** · August 8th 2010, 11:55 AM

Originally Posted by ragnar

[snip]
I looked at the link, Mr. Fantastic, but the part which gives the value of the sum just sort of states the result and I don't get the argument.
[snip]

Did you read ALL of the article in the link? The sum is done six different ways, the article certainly does not just "sort of states the result".

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