$$\sum _{k=0}^{\infty } \frac{k}{k^4+k^2+1}$$
Wolfram Alpha/Mathematica gives the precise answer in terms of a gamma function, but the numeric answer seems to be $\frac{1}{2}$. Is there any actual way to manipulate and telescope this sum somehow to arrive at that answer?
First note that $k^4+k^2+1 = (k^2+1)^2 - k^2 = (k^2+k+1)(k^2-k+1)$. Hence, $$\dfrac{k}{k^4 + k^2 + 1} = \dfrac12\left(\dfrac1{k^2-k+1} - \dfrac1{k^2+k+1}\right) = \dfrac12 \left(\dfrac1{k(k-1)+1} - \dfrac1{(k+1)k+1}\right)$$ Now telescope to get the answer. Using the above, in general, we have $$\sum_{k=0}^{N} \dfrac{k}{k^4 + k^2 + 1} = \dfrac12 - \dfrac12\dfrac1{N^2+N+1}$$ Taking the limit as $N \to \infty$, we get $\dfrac12$.