How to evaluate the sum : $\sum_{k=1}^{n} \frac{k}{k^4+1/4}$

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 $n\geq 1$.

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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.

Try to break the denominator into product of two factors:

$$\begin{align} 4k^4 + 1 &= (2k^2)^2 + 1 + 2 (2k^2) - 2 (2k^2) \\ &= (2k^2 +1)^2 - (2k)^2 \\ &= (2k^2 +2k +1)(2k^2 -2k+1) \end{align}$$

Using this we see the general term as:

$$T_k = \dfrac{1}{2k^2-2k+1} - \dfrac{1}{2k^2+2k+1} \\ T_{k+1} = \dfrac{1}{2k^2+2k+1} - \dfrac{1}{2(k+1)^2+2(k+1)+1} $$

Alternate terms cancel and the sum telescopes to:

$$1-\frac{1}{2n^2+2n+1}$$