Find the sum of the series $S = \sum_{k=1}^{n} \frac{k}{k^{4} + k^{2} + 1} $ $$ S = \sum_{k=1}^{n} \frac{k}{k^{4} + k^{2} + 1} $$
I started by factorizing the denominator as $k^2+k+1$ and $k^2-k+1$
The numerator leaves a quadratic with $k$ and $k-1$ or a constant with $k+1$ and $k-1.$
I tried writing the individual terms, ofcourse, it was useless.
How do I do this? 

So your term is equal to $$\frac{1}{2}\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)$$
Now note $(k+1)^2-(k+1)+1=k^2+k+1$, so your term is: $$\frac{1}{2}\left(\frac{1}{k^2-k+1}-\frac{1}{(k+1)^2-(k+1)+1}\right)$$
and you can apply a telescoping series technique to establish the sum to $n$ is just half of $\frac{1}{1^2 -1 +1}-\frac{1}{(n+1)^2-(n+1)+1}$. And if you are also looking for the limit, half of $\frac{1}{1^2 -1 +1}$.
