<s>[INST] 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.
$=\frac{1}{3}+\frac{3}{14}+\frac{2}{13}+0-\frac{1}{6}-\frac{1}{7}$
 [/INST] An observation:
We can reduce this series to another one. Note that $$\frac{1}{n(n\pm 1)+1}=\frac{|n\pm 1 -n|}{n(n\pm 1)+1}=|\tan(\arctan (n\pm 1)-\arctan n)|$$ Thus given summation is $$S=\sum_{n=1}^{50}\frac{1}{2n}\left(\tan(\arctan(n)-\arctan(n-1))-\tan(\arctan(n+1)-\arctan  n)\right)\\=\sum_{n=1}^{50}\frac{1}{2n}\left(\tan(\beta_n-\beta_{n-1})-\tan(\beta_{n+1}-\beta_n)\right)$$ where $\beta_n:=\arctan n,\ n\ge 1$. I am trying to think if there is a way to find this summation.
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