Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$ Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$
$$\begin{align}\frac{1}{n^4+n^2+1}&
=\frac{1}{n^4+2n^2+1-n^2}\\
&=\frac{1}{(n^2+1)^2-n^2}\\
&=\frac{1}{(n^2+n+1)(n^2-n+1)}\\
&=\frac{1-n}{2(n^2-n+1)}+\frac{1+n}{2(n^2+n+1)}\\
\end{align}$$
For $n={1,2,3}$  it is not giving telescooping series.
$=\frac{1}{3}+\frac{3}{14}+\frac{2}{13}+0-\frac{1}{6}-\frac{1}{7}$

Partial answer.
Rewrite the terms as:
$$\frac{1}{2}\left(\frac{n+1}{n^2+n+1} - \frac{n-1}{1+(n-1)+(n-1)^2}\right) = \\\frac{1}{2}\left(\frac{1}{n^2+n+1}+f(n)- f(n-1)\right)$$ where $f(n)=\frac{n}{1+n+n^2}$. In particular $\sum_{n=1}^\infty \left(f(n)-f(n-1)\right)=-f(0)=0$.
So we've reduced it to:
$$\frac12\sum \frac{1}{n^2+n+1}$$
Not sure how to compute that sum.
