How do I calculate $\sum_{n\geq1}\frac{1}{n^4+1}$? How do I calculate the following sum $$\sum_{n\geq1}\frac{1}{n^4+1}$$
 A: 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
$$\sum_{n=0}^{\infty}\frac{1}{n^2+a^2} =\frac{\pi}{a} \frac{e^{2a\pi}}{e^{2a\pi}-1} \,. $$
A: $$\frac{1}{x+1}=\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}+\frac{1}{x^5}..$$
$$\sum_{n=1}^\infty \frac{1}{n^4+1}=\zeta(4)-\zeta(8)+\zeta(12)-\zeta(16)+\zeta(20)...$$
$$\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2} = \zeta(2) t^2 - \zeta(4) t^4 + \zeta(6) t^6 - \zeta(8)t^8+\zeta(10)t^{10}\cdots.$$
$$\frac{\pi ti}{e^{2\pi t i}-1}-\frac{1}{2}+\frac{\pi t i}{2} = -\zeta(2) t^2 - \zeta(4) t^4 -\zeta(6) t^6 -\zeta(8)t^8-\zeta(10)t^{10} \cdots.$$
$$\frac{\pi ti}{e^{2\pi t i}-1}-\frac{1}{2}+\frac{\pi t i}{2} +\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2}=-2( \zeta(4)t^4+\zeta(8)t^8+\zeta(12)t^{12}+\zeta(16)t^{16}..)$$
I think you can figure out the rest...
A: 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?
A: 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.
A: 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 $b=\coth\frac{\pi}{\sqrt{2}}$.
