Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Related (but I don't think it's a duplicate): math.stackexchange.com/questions/141470/… –  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
add comment

3 Answers

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.

share|improve this answer
1  
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
1  
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
3  
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
show 4 more comments

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

share|improve this answer
add comment

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.

share|improve this answer
    
How are the sums geometric? –  Ishan Banerjee Feb 27 '13 at 4:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.