Evaluating $ \sum\frac{1}{1+n^2+n^4} $ How to evaluate following expression?
$$ \sum_{n=1}^{\infty}\frac{1}{1+n^2+n^4}$$
I doubt it is a telescopic Sum.
 A: Let us use the Abel-Plana formula
$$
\sum_{n=0}^\infty f(n)= \int_0^\infty f(x) \, dx+ \frac 1 2 f(0)+i \int_0^\infty \frac{f(i t)-f(-i t)}{e^{2\pi t}-1} \, dt
$$
for $f(n)=\dfrac{1}{1+n^2+n^4}$.
Since
$
f(it)=f(-it)
$
we obtain
$$
\sum_{n=0}^\infty f(n)= \int_0^\infty f(x) \, dx+ \frac 1 2 f(0).
$$
By standard way we get
$$
\int_0^\infty\frac{1}{1+x^2+x^4}=\frac{\pi \sqrt{3}}{6}.
$$
Now
$$
\sum_{i=1}^\infty f(n)=\sum_{i=0}^\infty f(n)-f(0)=\frac{\pi \sqrt{3}}{6}-\frac{1}{2}.
$$
A: Hint: Apply residue calculus on $$f(z)= \frac{\pi}{(1+z^{2}+z^{4})\tan(\pi z)}$$
Where you want to integrate over the square $S_{R}$ with vertices $(R+\frac{1}{2})(\pm1 \pm i)$
The reason is that $$Res(f(z), n\pi) = \frac{1}{1+n^{2}+n^{4}}$$
I'll leave the details to you!
A: Lots of nice analytical ideas! I will be trying to use them from now one. Numerically, it is not too difficult to get a nice answer. We use the first term of the Euler Maclaurin Summation formula which says:
$$\sum _{n=a}^b f(n)\sim \int_a^b f(x) \, dx+\frac{1}{2} (f(a)+f(b))$$
But it is alot more useful in getting the tail so we first compute
$$ \sum _{n=1}^{99} \frac{1}{n^4+n^2+1}=0.39907330193518 $$
for the integral on the RHS we use the trick that as $x$ gets large the $x^4$ term will completely drown out the $x^2 + 1$ terms.
$$\int_{100}^{\infty } \frac{1}{x^4+x^2+1} \, dx\approx \
\int_{100}^{\infty } \frac{1}{x^4} \, dx=\frac{1}{3000000}$$
Now it is just arithmetic,
$$0.39907330193518\, +\frac{1}{3000000}+\frac{1}{2} \left(\frac{1}{100^4+100^2+1}+0\right)$$
which equals $ 0.39907364026801 $ which is quite close, differing from the exact answer in the 11th place.
A: Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then:
$$ n^4+n^2+1 = (n^2-\omega)(n^2-\omega^2), $$
so:
$$\frac{1}{1+n^2+n^4}=\frac{1}{i\sqrt{3}}\left(\frac{1}{n^2-\omega}-\frac{1}{n^2-\omega^2}\right)$$
and:
$$\sum_{n=1}^{+\infty}\frac{1}{1+n^2+n^4}=\frac{1}{\sqrt{3}}\Im\sum_{n=1}^{+\infty}\frac{1}{n^2-\omega}$$
can be computed through the identity:
$$\sum_{n=1}^{+\infty}\frac{1}{n^2+a}=\frac{-1+\pi\sqrt{a}\coth(\pi\sqrt{a})}{2a}$$
that follows from considering the logarithmic derivative of the Weierstrass product for the $\sinh$ function. By putting all together, we have:

$$\sum_{n=1}^{+\infty}\frac{1}{1+n^2+n^4}=\frac{1}{6}\left(-3+\pi\sqrt{3}\tanh\frac{\pi\sqrt{3}}{2}\right)$$

as stated by WA.
A: First, we have
$$
\begin{align}
\frac1{z^4+z^2+1}
&=\frac1{12}\left(
\frac{-3-i\sqrt3}{z-e^{\pi i/3}}
+\frac{3+i\sqrt3}{z-e^{4\pi i/3}}
+\frac{3-i\sqrt3}{z-e^{2\pi i/3}}
+\frac{-3+i\sqrt3}{z-e^{5\pi i/3}}
\right)\tag{1}
\end{align}
$$
Let $\gamma$ be the rectangle
$$
[-1-i,1-i]\cup[1-i,1+i]\cup[1+i,-1+i]\cup[-1+i,-1-i]\tag{2}
$$
then the integral
$$
\frac1{2\pi i}\int_{(n+\frac12)\gamma}\frac{\pi\cot(\pi z)}{z^4+z^2+1}\,\mathrm{d}z\tag{3}
$$
tends to $0$ since along the horizontal paths, $|\pi\cot(\pi z)|\to\pi$ and along the vertical paths, $|\pi\cot(\pi z)|\lt\pi$.
Since $\pi\cot(\pi z)$ has residue $1$ at each integer, we get that $(3)$ is the sum of the residues of $(1)$ times $\pi\cot(\pi z)$ at the singularities of $(1)$ plus
$$
1+2\sum_{n=1}^\infty\frac1{n^4+n^2+1}\tag{4}
$$
The sum of the residues of $(1)$ times $\pi\cot(\pi z)$ at the singularities of $(1)$ is
$$
-4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\cot\left(\pi\tfrac{1+i\sqrt3}2\right)\right)\tag{5}
$$
Since the sum of $(4)$ and $(5)$ is $0$, we have
$$
\begin{align}
1+2\sum_{n=1}^\infty\frac1{n^4+n^2+1}
&=4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\cot\left(\pi\tfrac{1+i\sqrt3}2\right)\right)\\
&=-4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\tan\left(\pi\tfrac{i\sqrt3}2\right)\right)\\
&=\tfrac{\pi\sqrt3}3\tanh\left(\pi\tfrac{\sqrt3}2\right)\tag{6}
\end{align}
$$
Therefore,
$$
\sum_{n=1}^\infty\frac1{n^4+n^2+1}=\frac{\pi\sqrt3}6\tanh\left(\pi\frac{\sqrt3}2\right)-\frac12\tag{7}
$$
