<s>[INST] Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$ I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.

The original question was

Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)}$

I was able to simplify it to the following:
$\dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)} = \dfrac{(r^4 + r^3 - r^2 - r)}{(r^4 + r^2 + 1)(r^2 + r)} + \dfrac{3r^2+r+1}{\{(r+1)(r^2 + r + 1)\}\{r(r^2 - r + 1)\}} = \dfrac{r^2 - 1}{r^4 + r^2 + 1} + \left[\dfrac{1}{r(r^2 - r + 1)} - \dfrac{1}{(r+1)(r^2 + r + 1)}\right]$
The second term can be evaluated using telescoping, and the first term is what this post is asking for.
Any other ways of solving the original question are also welcome.
 [/INST] Write \begin{align}&\frac{r^2-1}{r^4+r^2+1} = \frac{Ar+B}{r^2-r+1}+\frac{Cr+D}{r^2+r+1}\\ \iff &r^2-1 = (Ar+B)(r^2+r+1)+(Cr+D)(r^2-r+1)\tag{1}\end{align}
You can now expand the RHS as a polynomial in $r$ and by direct comparison of coefficients on the LHS and the RHS get a $4\times4$ linear system.
Alternatively, let $\alpha$ be a root of $x^2-x+1$. In particular, we have $\alpha^2 = \alpha - 1$. Evaluating $(1)$ at $\alpha$ gives us
\begin{align}\alpha^2-1 &= (A\alpha + B)(\alpha^2+\alpha+1)\\
(\alpha-1)-1 &= (A\alpha + B)((\alpha-1)+\alpha+1)\\
\alpha - 2 &= 2\alpha(A\alpha+B)\\
\alpha - 2 &= 2A\alpha^2+2B\alpha\\
\alpha - 2 &= 2A(\alpha - 1)+2B\alpha\\
\alpha - 2 &= 2(A+B)\alpha-2A \implies 2(A+B)=1,\ -2A = -2
\end{align}
Note that it's important to get rid of any higher powers of $\alpha$ besides $\alpha$ and $1$ before you compare coefficients. Similarly, if you take $\beta$ a root of $x^2+x+1$, you can calculate $C$ and $D$ analogously.
Now that you can calculate partial fractions decomposition, we have
\begin{align}
\sum_{r = 1}^n \frac{r^2-1}{r^4+r^2+1} &=\sum_{r = 1}^n\left( \frac{2r-1}{2(r^2-r+1)}-\frac{2r+1}{2(r^2+r+1)} \right) \\
&= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 1}^n \frac{2r+1}{2(r^2+r+1)}\\
&= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 1}^n \frac{2(r+1)-1}{2((r+1)^2-(r+1)+1)} \\
&= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 2}^{n+1} \frac{2r-1}{2(r^2-r+1)} \\
&= \frac 12 - \frac{2n+1}{2(n^2+n+1)}
\end{align}
Taking a limit as $n\to \infty$ gives us $$\sum_{r = 1}^\infty \frac{r^2-1}{r^4+r^2+1} =  \frac 12.$$
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