TITLE: Evaluating $ \sum\frac{1}{1+n^2+n^4} $
QUESTION [12 upvotes]: How to evaluate following expression?
$$ \sum_{n=1}^{\infty}\frac{1}{1+n^2+n^4}$$
I doubt it is a telescopic Sum.

REPLY [4 votes]: 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.