# Hard sum question?

1. Apr 8, 2010

### ~Death~

What is sum n=1 to infinity n^2/(n^4+1)

2. Apr 9, 2010

### CRGreathouse

I can calculate it quite precisely -- the zeta function can accelerate it as much as desired -- but I don't have a nice expression for it.

3. Apr 9, 2010

### Gib Z

$$\sum_{n=0}^{\infty} \frac{n^2}{1+n^4} = \frac{1+i}{4\sqrt{2}} \pi \left(i \cot \left(\frac{1+i}{\sqrt{2}} \pi\right) - \cot \left(\frac{1-i}{\sqrt{2}} \pi\right )\right) \approx 1.12852792472431...$$

4. Apr 9, 2010

### ~Death~

ahh
that looks scarry but thanks
any insight on how you got it?
I was thinking residues but they look really ugly

5. Apr 9, 2010

### Gib Z

It all came to me in a dream...

...In the dream, I was on my laptop and typing the sum into Mathematicia.

lol sorry I can't see any good line of attack for this problem. Sometimes when I'm stuck for ideas I plug in the expression into Mathematicia (or the online version WolframAlpha.com if you don't have it) and try to get some ideas from what the answer looks like, what functions it uses, etc etc. But this time, The only idea I got from the Cotangents and roots of -1 were residues as well, which I realized turned very ugly very quickly, as you did.

6. Apr 10, 2010

### Count Iblis

This is actually a standard complex analysis exam problem. Summation over integers is summing over the zeroes of the function sin(pi z). If you want to sum over the zeroes of a function f(z), you may want to look at the function f'/f, as this function has its poles at the zeroes of f with residue 1. So, the contour integral:

$$\frac{1}{2\pi i}\oint \frac{f'(z)}{f(z)} g(z) dz$$

will be equal to the summation of g(z) over the zeroes of f that are inside the contour. If f has poles then these will contribute also to the summation of g but with a minus sign. In addition to the poles of f'/f the function g(z) may have poles itself which will contribute to the integral.

Now, what if the function g is such that the contour integral over a coircle of radius R tends to zero in the limit R to infinity? Then the summation of g over the zeroes of f will be determined by the residues of f'/f g at the poles of g.

7. Apr 11, 2010

### uart

Thanks for the explaination Count, that's a nice way to do it.

So the sum can be written as :

$$-\frac{1}{2} \sum_{k=1}^4 \pi \cot(\pi a_k) \lim_{x \rightarrow a_k} \frac{x^2 (x-a_k) }{x^4 + 1}$$

Where the four $a_k$'s are the complexs roots of x^4+1=0.

BTW. Just in case anyone is wondering about the 1/2 out the front of that sum. It's because the contour of integration runs a full circle (at infinity) around the complex plane and so includes the sum over negative integers as well.