How to evaluate the sum : $\sum_{k=1}^{n} \frac{k}{k^4+1/4}$ I have been trying to figure out how to evaluate the following sum:
$$S_n=\sum_{k=1}^{n} \frac{k}{k^4+1/4}$$
In the problem, the value of $S_{10}$ was given as $\frac{220}{221}$. 
I have tried  partial decomposition, no where I go. Series only seems like it telescopes, otherwise there isn't another way.
Any ideas are appreciated! 

By Sophie Germain's identity
$$ 4k^4+1 = (2k^2+2k+1)(2k^2-2k+1) \tag{A}$$
hence
$$ \frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1} = \frac{4k}{4k^4+1} = \frac{k}{k^4+1/4}\tag{B} $$
and we may notice that by setting $p(x)=2x^2-2x+1$ we have $p(x+1)=2x^2+2x+1$.
In particular
$$ \sum_{k=1}^{n}\frac{k}{k^4+1/4}=\sum_{k=1}^{n}\left(\frac{1}{p(k)}-\frac{1}{p(k+1)}\right) = \frac{1}{p(1)}-\frac{1}{p(n+1)}=1-\frac{1}{2n^2+2n+1} $$
equals $\frac{2n^2+2n}{2n^2+2n+1}$ for any $n\geq 1$.

Telescoping is not strictly necessary to be able to compute the value of similar series. For instance
$$ \sum_{k\geq 0}\frac{1}{k^4+4} = \frac{\pi\cos\pi+\sinh\pi}{8\sinh\pi}, $$
but this is a different story, related with Weierstrass products, the Poisson summation formula or the (inverse) Laplace transform.
