# Thread: Sum (kind of difficult)

1. ## Sum (kind of difficult)

Compute $\displaystyle \sum_{n=0}^{\infty}\frac{1}{\pi-en^2}$

And if you cannot find it see how closely you can bound it.

2. Originally Posted by Mathstud28
Compute $\displaystyle \sum_{n=0}^{\infty}\frac{1}{\pi-en^2}$

And if you cannot find it see how closely you can bound it.
Let $\displaystyle \mathcal{F}(x)$ be function such that $\displaystyle \forall{x}\in\left[-\delta,\delta\right]~\mathcal{F}(x)=\sin^2(ax)~a\notin\mathbb{Z}$. Furthermore let $\displaystyle \mathcal{F}(x)$ have the charcteristic that $\displaystyle \mathcal{F}(x)\in{P}_{2\delta}\implies\mathcal{F}\ left(x+2\delta\right)=\mathcal{F}(x)~\forall{x}\in \mathbb{R}$. Now it is pretty clear that $\displaystyle \mathcal{F}(x)\in\mathcal{C}\left(\mathbb{R}\right )$ except possibly at $\displaystyle \delta\mathbb{Z}$, but in fact $\displaystyle \mathcal{F}(x)$ is continuous at all these points due to its eveness. Next note that $\displaystyle \mathcal{F}(x)\in{D^1}$, in other words its posses a right and left hand limit everywhere. So since noting all of these things we may conclude that $\displaystyle S_n(x)\to\mathcal{F}(x)~\forall{x}\in\mathbb{R}$ where $\displaystyle S_n(x)$ is the Fourier Series given by.

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

Where

\displaystyle \begin{aligned}A_0&=\frac{1}{\delta}\int_{-\delta}^{\delta}\mathcal{F}(x)dx\\ &=\frac{1}{\delta}\int_{-\delta}^{\delta}\sin^2(ax)dx\\ &=\frac{2a\delta-\sin\left(2a\delta\right)}{2a\delta} \end{aligned}

\displaystyle \begin{aligned}A_n&=\frac{1}{\delta}\int_{-\delta}^{\delta}\mathcal{F}(x)\cos\left(\frac{n\pi {x}}{\delta}\right)dx\\ &=\frac{1}{\delta}\int_{-\delta}^{\delta}\sin^2(ax)\cos\left(\frac{n\pi{x}} {\delta}\right)dx\\ &=-2a\delta\sin\left(2a\delta\right)\frac{(-1)^n}{(2a\delta)^2-n^2\pi^2}~\forall{n\in\mathbb{N}} \end{aligned}

\displaystyle \begin{aligned}B_n&=\frac{1}{\delta}\int_{-\delta}^{\delta}\mathcal{F}(x)\sin\left(\frac{n\pi {x}}{\delta}\right)dx\\ &=\frac{1}{\delta}\int_{-\delta}^{\delta}\sin^2(ax)\sin\left(\frac{n\pi{x}} {\delta}\right)dx\\ &=0~\because\sin^2(ax)\sin\left(\frac{n\pi{x}}{\de lta}\right)\text{ is odd} \end{aligned}

$\displaystyle \therefore~S_n(x)=\frac{2a\delta-\sin\left(4a\delta\right)}{2a\delta}-2a\delta\sin\left(2a\delta\right)\sum_{n=1}^{\inft y}\frac{(-1)^n}{\left(2a\delta\right)^2-n^2\pi^2}\cos\left(\frac{n\pi{x}}{\delta}\right)$

And as was stated above $\displaystyle S_n(x)\to\mathcal{F}(x)$

$\displaystyle \therefore~\mathcal{F}(x)=\frac{2a\delta-\sin\left(2a\delta\right)}{4a\delta}-2a\delta\sin\left(2a\delta\right)\sum_{n=1}^{\inft y}\frac{(-1)^n}{\left(2a\delta\right)^2-n^2\pi^2}\cos\left(\frac{n\pi{x}}{\delta}\right)$

So $\displaystyle \forall{x}\in\left[-\delta,\delta\right]$

$\displaystyle \sin^2(ax)=\frac{2a\delta-\sin\left(2a\delta\right)}{4a\delta}-2a\delta\sin\left(2a\delta\right)\sum_{n=1}^{\inft y}\frac{(-1)^n}{\left(2a\delta\right)^2-n^2\pi^2}\cos\left(\frac{n\pi{x}}{\delta}\right)$

Now letting $\displaystyle 2a=z$ gives

$\displaystyle \sin^2\left(\frac{zx}{2}\right)=\frac{z\delta-\sin\left(z\delta\right)}{2z\delta}-z\delta\sin\left(z\delta\right)\sum_{n=1}^{\infty} \frac{(-1)^n}{\left(z\delta\right)^2-n^2\pi^2}\cos\left(\frac{n\pi{x}}{\delta}\right)$

So now consider when $\displaystyle x=\delta=\pi$ we are left with

$\displaystyle \sin^2\left(\frac{z\pi}{2}\right)=\frac{z\pi-\sin(z\pi)}{2z\pi}-z\pi\sin\left(z\pi\right)\sum_{n=1}^{\infty}\frac{ 1}{z^2\pi^2-n^2\pi^2}$

Solving gives

$\displaystyle \forall{z}\notin2\mathbb{Z}~\frac{-\pi\sin^2\left(\frac{z\pi}{2}\right)^2}{z\sin(z\pi )}+\frac{\pi}{2z\sin(z\pi)}-\frac{1}{2z^2}=\sum_{n=1}^{\infty}\frac{1}{z^2-n^2}$

So then we get finally that

$\displaystyle \forall{z}\notin2\mathbb{Z}~\frac{-\pi\sin^2\left(\frac{{z}\pi}{2}\right)^2}{z\sin(z\ pi)}+\frac{\pi}{2z\sin(z\pi)}+\frac{1}{2z^2}=\sum_ {n=0}^{\infty}\frac{1}{z^2-n^2}$

$\displaystyle \boxed{\therefore~\sum_{n=0}^{\infty}\frac{1}{\pi-en^2}=\frac{\sin^2\left(\frac{\pi^{\frac{3}{2}}}{2 \sqrt{e}}\right)\sqrt{\pi}}{\sqrt{e}\sin\left(\fra c{\pi^{\frac{3}{2}}}{\sqrt{e}}\right)}-\frac{\sqrt{\pi}}{2\sqrt{e}\sin\left(\frac{\pi^{\f rac{3}{2}}}{\sqrt{e}}\right)}-\frac{1}{2\pi}}$

Note from this we can also glean that

$\displaystyle \forall{z}\notin2\mathbb{Z}~\sum_{n=0}^{\infty}\fr ac{(-1)^n}{z^2-n^2}=\frac{\pi}{2z\sin(z\pi)}+\frac{1}{2z^2}$

3. Originally Posted by PaulRS
Because it's not very different from the others you've posted.
Hey! Don't be a Negative Nancy! Haha, your right though
We have seen that: $\displaystyle \sum\limits_{n = - \infty }^{ + \infty } {\tfrac{1} {{n^2 + x^2 }}} = \tfrac{{\pi \cdot \coth \left( {\pi \cdot x} \right)}} {x}$ which actually holds whenever $\displaystyle x \in \mathbb{C}$ and $\displaystyle n^2 + x^2 \ne 0{\text{ }}\forall n \in \mathbb{Z}$

Change $\displaystyle x \to i \cdot x$ and then set $\displaystyle x = \sqrt {\tfrac{\pi } {e}}$

See the Herglotz trick
I like this solution because it is simplistic and powerful and in so beautiful. I dislike it because it doesn't take much thought. Picking a Fourier Series is much harder than one would imagine. Also, it can become repetitive, not that my Fourier Series are that much different each time though...and FS looks cooler Regardless, thanks again Paul for a nice solution.