# How do I calculate $\sum_{n\geq1}\frac{1}{n^4+1}$?

How do I calculate the following sum $$\sum_{n\geq1}\frac{1}{n^4+1}$$

• ok :) I have calculated the sum partial. Right? – Sophie Germain Feb 16 '13 at 22:26
• $$\sum_{n=1}^\infty \frac{1}{n^4+1}=-\frac{1}{4}(-2+(-1)^{1/4}\pi \cot((-1)^{1/4}\pi)+(-1)^{3/4}\pi \cot((-1)^{3/4}\pi))$$ – Ethan Feb 16 '13 at 22:29
• @julien a root of unity – Ethan Feb 16 '13 at 22:41
• @Ethan Which one? Does it mean your forumla is the same no matter what fourth root of $-1$ you choose? – Julien Feb 16 '13 at 22:43

Hint Consider the function $f(z) = \dfrac{\pi \cot \pi z}{z^4+1}$. I assume you have seen other examples of computing series with residue calculus (since you marked the question complex-analysis).

A few more details. Let $C(z) = \pi\cot \pi z$. Then $C$ is holomorphic everywhere except at the integers, and $\newcommand{\Res}{\operatorname{Res}}\Res(C;z=n) = 1$ for every $n \in \mathbb{Z}$. The function $f$ has additional simple poles at points where $z^4+1=0$.

Let $\gamma_N$ be the positively oriented square with corners at $\pm(N+\frac12) \pm(N+\frac12)i$. A tedious computation will show that $|C(z)|$ is bounded on $\gamma_N$ by a constant, not depending on $N$. (It you haven't seen this, try to work out the details yourself. Check a textbook if you can't get it.)

The residue theorem gives

$$\int_{\gamma_N} f(z)\,dz = \sum_{k=-N}^N \Res(f;z=k) + \sum_{\alpha^4=-1} \Res(f;z=\alpha).$$

Letting $N\to\infty$, we end up with (since $|f|\to0$ quickly enough at $\infty$):

$$0 = \sum_{k=-\infty}^\infty \frac{1}{k^4+1} + \sum_{\alpha^4=-1} \Res(f;z=\alpha).$$

What remains is to compute the four extra residues, and manipulate the doubly infinite series a little, but since it's homework, I'm not going to finish things off for you.

• thanx ;) I will do it – Sophie Germain Feb 16 '13 at 22:47
• I can't with this exercise, please help me. – Sophie Germain Feb 17 '13 at 1:50
• ok, thanx :) it's very easy. – Sophie Germain Feb 17 '13 at 17:13

First, we write the sum in the following form

$$\sum_{n=0}^{\infty}\frac{1}{n^4+1} = \frac{i}{2}\sum_{n=0}^{\infty}\frac{1}{n^2+i} -\frac{i}{2} \sum_{n=0}^{\infty}\frac{1}{n^2-i}\quad i=\sqrt{-1},$$

Then, we use the result

$$\sum_{n=0}^{\infty}\frac{1}{n^2+a^2} =\frac{\pi}{a} \frac{e^{2a\pi}}{e^{2a\pi}-1} \,.$$

$$\frac{1}{x+1}=\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}+\frac{1}{x^5}..$$ $$\sum_{n=1}^\infty \frac{1}{n^4+1}=\zeta(4)-\zeta(8)+\zeta(12)-\zeta(16)+\zeta(20)...$$ $$\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2} = \zeta(2) t^2 - \zeta(4) t^4 + \zeta(6) t^6 - \zeta(8)t^8+\zeta(10)t^{10}\cdots.$$ $$\frac{\pi ti}{e^{2\pi t i}-1}-\frac{1}{2}+\frac{\pi t i}{2} = -\zeta(2) t^2 - \zeta(4) t^4 -\zeta(6) t^6 -\zeta(8)t^8-\zeta(10)t^{10} \cdots.$$ $$\frac{\pi ti}{e^{2\pi t i}-1}-\frac{1}{2}+\frac{\pi t i}{2} +\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2}=-2( \zeta(4)t^4+\zeta(8)t^8+\zeta(12)t^{12}+\zeta(16)t^{16}..)$$

I think you can figure out the rest...

• i loved wolfram alpha too but I need the development of the exercise. – Sophie Germain Feb 17 '13 at 1:48
• @SophieGermain Are you referring to my previous comment? – Ethan Feb 17 '13 at 3:58
• @SophieGermain: Ethan's answer here can be turned into a full solution. Let $$f(t)=\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2},$$ and then consider $$f(\zeta_8 t)+f(\zeta_8^3t)$$ where $\zeta_8=e^{\pi i/8}$ is an eight root of unity. Now, to get the series $\sum_{n=1}^\infty \frac{1}{n^4+1}$, you need to take the limit as $t\rightarrow 1$, and use Abel's limit theorem to justify the switching of the orders. – Eric Naslund Feb 17 '13 at 16:43

I think that:

If we know $$\sum_{n\geq1}\frac{1}{n^4-a^4} = \frac{1}{2a^2}-\frac{\pi}{4 a^3}(\cot \pi a+\coth \pi a)$$ Then, with $a = \sqrt[4]{-1}$: $$\sum_{n\geq1}\frac{1}{n^4+1} \approx 0.57847757966713683831802219...$$ Right?

It is enough to apply the Poisson summation formula. Since the Cauchy distribution and the Laplace distribution are conjugated via the Fourier transform, a simple partial fraction decomposition ensures

$$\mathscr{F}\left(\frac{1}{1+x^4}\right)(s) =\frac{\pi}{\sqrt{2}}e^{-\pi\sqrt{2}|s|}(\cos+\sin)(\pi\sqrt{2}|s|)$$ hence by summing the real and imaginary part of a geometric series $$\sum_{n\in\mathbb{Z}}\frac{1}{1+n^4}=\frac{\pi}{\sqrt{2}}\cdot\frac{(b-a)+ab(a+b)}{a^2+b^2}$$ where $$a=\cot\frac{\pi}{\sqrt{2}}$$ and $$b=\coth\frac{\pi}{\sqrt{2}}$$.