1. ## A sum

Evaluate $\sum_{k=0}^\infty\frac1{(2k+1)^2}.$

Cheers

2. Originally Posted by Krizalid
Evaluate $\sum_{k=0}^\infty\frac1{(2k+1)^2}.$

Cheers
Solution 1: Let $f(z) = \frac{1}{(2z+1)^2}$ it has a singularity at $z=-1/2$. That means $\sum_{n\in \mathbb{Z}} \frac{1}{(2n+1)^2} = - \mbox{res}\left( \pi f(z) \cot\pi z, -1/2 \right)$.

The residue is $\lim_{z\to -1/2} \left( \frac{\pi \cot \pi z(z+1/2)^2}{(2z+1)^2}\right)' = \frac{\pi}{4} \lim_{z\to -1/2} \left( \cot \pi z \right)' = - \frac{\pi^2}{4} \csc^2 \pi z$.

So the sum is (if I did not make any computational mistakes) $\frac{\pi^2}{8}$.

3. Solution 2: We know $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$. This means $\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n)^2} = \frac{\pi^2}{6}$.
But $\sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{24}$. This means $\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{\pi^2}{8}$.

4. Okay.

Here's a solution that I really love:

$\sum\limits_{k = 0}^\infty {\frac{1}
{{(2k + 1)^2 }}} = \int_0^1 {\int_0^1 {\left\{ {\sum\limits_{k = 0}^\infty {\left( {x^2 y^2 } \right)^k } } \right\}\,dx} \,dy} = \int_0^1 {\int_0^1 {\frac{{dx\,dy}}
{{1 - x^2 y^2 }}} } .$

Substitute $(x,y) = \left( {\frac{{\sin u}}
{{\cos v}},\frac{{\sin v}}
{{\cos u}}} \right).$
The Jacobian matrix is $1-x^2y^2,$ and the integral becomes

$\iint_A {du\,dv},$ where $A$ is the triangle with coordinates $(0,0),\,\bigg( {\frac{\pi }
{2},0} \bigg),\,\bigg( {0,\frac{\pi }
{2}} \bigg).$
Finally

\begin{aligned}
\sum\limits_{k = 0}^\infty {\frac{1}
{{(2k + 1)^2 }}} &= \int_0^{\pi /2} {\int_0^{\pi /2 - v} {\,du} \,dv}\\
&= \int_0^{\pi /2} {\bigg( {\frac{\pi }
{2} - v} \bigg)\,dv}\\
&= \frac{{\pi ^2 }}
{8},
\end{aligned}

as required.

5. Haha, I know where you got that from. My professor used that too when we were doing the Jacobian.

6. Can this be solved using the Riemann Zeta function?

7. Originally Posted by colby2152
Can this be solved using the Riemann Zeta function?

8. Sorry I'm bored.

Consider some function $f$. Let $f(x)=|x|\quad\forall{x}\in[-\pi,\pi]$. Let the function also posses the charcteristic $f\left(x+2\pi\right)=f(x)\quad\forall{x}\in\mathbb {R}$. So now we seek a Fourier Series to describe this function. This series will be of the form

$f(x)=\frac{A_0}{2}+\sum_{n=1}^{\infty}\bigg[A_n\cos\left(\frac{n\pi{x}}{\delta}\right)+B_n\sin \left(\frac{n\pi{x}}{\delta}\right)\bigg]$

Where $\delta$ is a number such that $f\left(x+2\delta\right)=f(x)$

$A_0=\frac{1}{\delta}\int_{-\delta}^{\delta}f(x)dx$

$A_n=\frac{1}{\delta}\int_{-\delta}^{\delta}f(x)\cos\left(\frac{n\pi{x}}{\delt a}\right)dx$

$B_n=\frac{1}{\delta}\int_{-\delta}^{\delta}f(x)\sin\left(\frac{n\pi{x}}{\delt a}\right)dx$

So for $f(x)=|x|$ and $\delta=\pi$ we have

$f(x)=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|dx+\sum_{n=1}^{\infty}\bigg[\frac{1}{\pi}\int_{-\pi}^{\pi}|x|\cos(nx)dx\cos(n{x})+\frac{1}{\pi}\in t_{-\pi}^{\pi}|x|\sin(nx)dx\sin(nx)\bigg]$

Now $\frac{1}{\pi}\int_{-\pi}^{\pi}|x|dx=\frac{\pi}{2}$

$\frac{1}{\pi}\int_{-\pi}^{\pi}|x|\cos(nx)dx$

$=\frac{2}{\pi}\cdot\frac{\cos(n\pi)-1}{n^2}$

${{=}}{{0\quad{n\text{ is even}}}\brace{\frac{-4}{\pi{n^2}}\quad{n}\text{ is odd}}}$

And $|x|\cdot{\sin(nx)}$ is odd which implies that

$B_n=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|\sin(nx)dx=0$

$\therefore|x|=\frac{\pi}{2}+\frac{2}{\pi}\sum_{n=0 }^{\infty}\frac{(-1)^n-1}{n^2}\cos(nx)$

$=\frac{\pi}{2}+\frac{2}{\pi}\sum_{n=1,3,5\cdots}^{ \infty}\frac{(-1)^n-1}{n^2}\cos(nx)+\frac{2}{\pi}\sum_{n=2,4,6\cdots}^ {\infty}\frac{(-1)^n-1}{n^2}\cos(nx)$

$=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1,3,5,\cdots}^{\infty}\frac{\ cos(nx)}{n^2}+\sum_{n=2,4,6,\cdots}^{\infty}0$

$=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{\cos(nx)}{(2 n+1)^2}$

Therefore we can see that

$f(0)=|0|=0=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$

Solving gives

$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2} {8}\quad\blacksquare$

I'm sure this is solution is common...I'm not sure though...I found this today while tinkering.

9. I'm sure this is solution is common...I'm not sure though...I found this today while tinkering.
Two or three teachers used Fourier series, last year, to show us that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$, so nothing that uncommon :P

10. Originally Posted by Moo
Two or three teachers used Fourier series, last year, to show us that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$, so nothing that uncommon :P
Yeah, I found that while tinkering too. I just found the fourier series for some functions and these presented themselves as particulars of these. I ended up calculating the even values $\zeta(x)$ up to 10. Anyway, I have always been more fan of comparing the coefficents of

$\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x\prod_{n=1}^{\infty}\bigg( 1-\frac{x^2}{n^2\pi^2}\bigg)$

11. Quoting MOO:

Two or three teachers used Fourier series, last year, to show us that , so nothing that uncommon :P