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So my question is to find the value of $\sum_{k=0}^\infty\frac{1}{k^2+1}$ and more generally $\sum_{k=0}^\infty\frac{1}{Q(k)}$ where Q is a quadratic polynomial with no zeroes on the integers.

I can prove it converges, by the comparison test. I think I 've seen such a sum before, and I think it has a hyperbolic function solution. But the way they did it there involved complex analysis, and I'm not very comfortable with that.

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Related (but I don't think it's a duplicate):… – SSumner Feb 26 '13 at 16:19
fix your sum, you have $k$, $n$... – Cortizol Feb 26 '13 at 16:34
Okay, for the 1st part of the question, I can use what SSumner wrote and write $\frac{1}{k^2+a^2}=\frac{1}{k^2-(ia)^2}$ – Ishan Banerjee Feb 26 '13 at 16:39
up vote 5 down vote accepted

Using the residue theorem, you can show that

$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2+a^2} = \frac{\pi}{a} \coth{\pi a}$$

This is equivalent to saying that

$$\sum_{n=1}^{\infty} \frac{1}{n^2+a^2} = \frac{1}{2} \left (\frac{\pi}{a} \coth{\pi a} - \frac{1}{a^2}\right )$$

You can also derive this by considering the Maclurin expansion of $z \coth{z}$:

$$z \coth{z} = 1 + \sum_{k=1}^{\infty} \frac{B_{2 k} (2 z)^{2 k}}{(2 k)!}$$

where $B_{2 k}$ is a Bernoulli number, which also shows up in Riemann zeta functions of even, positive argument:

$$\zeta(2 k) = (-1)^{k+1} \frac{B_{2 k} (2 \pi)^{2 k}}{2 (2 k)!}$$

To evaluate the sum, factor out $n^2$ from the denominator and Taylor expand:

$$\begin{align}\sum_{n=1}^{\infty} \frac{1}{n^2+a^2} &= \frac{1}{n^2} \frac{1}{1+ \frac{a^2}{n^2}}\\ &= \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{\infty} (-1)^k \left (\frac{a^2}{n^2}\right )^{k} \\ &=\sum_{k=0}^{\infty} (-1)^k a^{2 k}\sum_{n=1}^{\infty} \frac{1}{n^{2 k+2}} \\ &=\sum_{k=0}^{\infty} (-1)^k a^{2 k} \zeta(2 k+2)\\ &= \frac{1}{2 a^2}\sum_{k=1}^{\infty} \frac{B_{2 k} (2 \pi a)^{2 k}}{(2 k)!} \\ &= \frac{1}{2 a^2} ( \pi a \coth{\pi a} - 1)\\ \end{align}$$

The result follows.

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Nice, but I'd like to wait for more answers. Are there ways to generalize this? – Ishan Banerjee Feb 26 '13 at 16:47
What way did you have in mind? I could imagine using this to evaluate stuff like $$\sum_{n=1}^{\infty} \frac{1}{(n^2+a^2)^m}$$, but that could get messy. Another generalization could be something like $$\sum_{n=1}^{\infty} \frac{(-1)^n}{(n^2+a^2)}$$, which would be relatively easy, or $$\sum_{n=1}^{\infty} \frac{\cos{\beta n}}{(n^2+a^2)}$$ which could get tricky. – Ron Gordon Feb 26 '13 at 16:51
Nice,Do you use $f(z)=\frac{\pi \cot \pi z}{a^2+z^2}$ and integrating over $|z|=N+1/2$ and letting $N \rightarrow +\infty$ to get $\sum_{n=-\infty}^{\infty} \frac{1}{n^2+a^2} = \frac{\pi}{a} \coth{\pi a}$? – Laura Feb 26 '13 at 16:52
@Tai: for using residue theory to evaluate the sum yes, but the point was to not do that here. Rather, I used a Taylor series expansion to create a double sum which could be evaluated nicely. – Ron Gordon Feb 26 '13 at 16:55
Incidentally, the sum $\sum\dfrac{(-1)^n}{n^2+a^2}$ has a much easier approach once you know the baseline sum $S(a)=\sum\dfrac{1}{n^2+a^2}$: $\sum_n\dfrac{(-1)^n}{n^2+a^2}=2\sum_k\dfrac{1}{(2k)^2+a^2}-\sum_k\dfrac{1}{k^2+‌​a^2} =\frac12\sum_k\dfrac{1}{k^2+(a/2)^2}-\sum_k\dfrac{1}{k^2+a^2}=\frac12S(a/2)-S(a)‌​$. – Steven Stadnicki Feb 26 '13 at 19:07

Look at function $ f(x)=e^{ax},~-\pi\leqslant x\leqslant\pi $. It is known (evaluate) that $ f(x)=e^{ax}, ~-\pi\leqslant x\leqslant\pi $ has the Fourier series

$$ \frac{\sinh\pi a}{\pi a}+\frac{2\sinh\pi a}{\pi}\sum_{n=1}^{\infty}(-1)^n\left[\left(\frac{a}{a^2+n^2}\right)\cos{nx}-\left(\frac{n}{a^2+n^2}\right)\sin nx\right]. $$

The function $f(x)$ is continuous for $ -\pi\leqslant x\leqslant\pi $, but $ f(-\pi)\neq f(\pi) $. Thus, at the ends of the fundamental interval the Fourier series for $f(x)$ will converge to the value $$ \frac{1}{2}\left(f(\pi)+f(-\pi)\right)=\frac{1}{2}\left(e^{a\pi}+e^{-a\pi}\right)=\cosh\pi a. $$

Using this result and setting $x=\pi$ in the Fourier series gives $$\cosh\pi a =\frac{\sinh\pi a}{\pi a}+\frac{2\sinh\pi a}{\pi}\sum_{n=1}^{\infty}\left(\frac{a}{a^2+n^2}\right),$$

where use have been made of the result $ \cos n\pi = (-1)^n $. Thus $$ \coth\pi a =\frac{1}{\pi}\left[\frac{1}{a}+2\sum_{n=1}^{\infty}\frac{a}{a^2+n^2}\right]$$ or, equivalently $$\frac{1}{2}\left(\pi\coth\pi a-\frac{1}{a}\right) =\sum_{n=1}^{\infty}\frac{a}{a^2+n^2}.$$

And now we find $$\sum_{n=1}^{\infty} \frac{1}{n^2+a^2} = \frac{1}{2} \left (\frac{\pi}{a} \coth{\pi a} - \frac{1}{a^2}\right ).$$

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You can divide the summand into partial fractions (yes, you'll get some nice complex numbers in the process), and your sum splits into two geometric sums. Not as nice a form as the solution by rlgordonma, sorry.

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How are the sums geometric? – Ishan Banerjee Feb 27 '13 at 4:54

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