8. Compute
$$
\sum_{n=0}^{\infty} \frac{n}{n^{4}+n^{2}+1} .
$$
Solution:
$1 / 2$
Note that
$$
n^{4}+n^{2}+1=\left(n^{4}+2 n^{2}+1\right)-n^{2}=\left(n^{2}+1\right)^{2}-n^{2}=\left(n^{2}+n+1\right)\left(n^{2}-n+1\right) .
$$

Decomposing into partial fractions, we find that
$$
\frac{n}{n^{4}+n^{2}+1}=\frac{1}{2}\left(\frac{1}{n^{2}-n+1}-\frac{1}{n^{2}+n+1}\right) .
$$

Now, note that if $f(n)=\frac{1}{n^{2}-n+1}$, then $f(n+1)=\frac{1}{(n+1)^{2}-(n+1)+1}=\frac{1}{n^{2}+n+1}$. It follows that
$$
\sum_{n=0}^{\infty} \frac{n}{n^{4}+n^{2}+1}=\frac{1}{2}((f(0)-f(1))+(f(1)-f(2))+(f(2)-f(3))+\cdots) .
$$

Since $f(n)$ tends towards 0 as $n$ gets large, this sum telescopes to $f(0) / 2=1 / 2$.