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.
 A: If you are comfortable with generalized harmonic number, you could consider first the partial sum
$$S_n=\sum_{r=1}^{n} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$$ and write first
$$\dfrac{r^2 - 1}{r^4 + r^2 + 1}=\frac{(r-1)(r+1)}{(r-a)(r-b)(r-c)(r-d)}$$
Using partial fraction decomposition, this is
$$\frac{a^2-1}{(a-b) (a-c) (a-d) (r-a)}+\frac{b^2-1}{(b-a) (b-c)
   (b-d) (r-b)}+$$ $$\frac{c^2-1}{(c-a) (c-b) (c-d)
   (r-c)}+\frac{d^2-1}{(d-a) (d-b) (d-c) (r-d)}$$ and use
$$\sum_{r=1}^n \frac 1{r-k}=H_{n-k}-H_{-k}$$
After simplification
$$S_n=\frac{(n-1) n}{2 \left(n^2+n+1\right)}$$
A: 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 $$\sum_{r = 1}^\infty \frac{r^2-1}{r^4+r^2+1} =  \frac 12.$$
A: As this Art of Problem Solving thread suggests, write this using a  form of partial fractions:
$$(Ar + B)(r^2 - r  + 1) + (Cr + D)(r^2 + r + 1) = r^2 - 1$$
Now comparing the coefficients of $r^3$, $r^2$, $r$, and the constant term, we end up with $A + C = 0, B - A + D + C = 1, -B + A+D+C=0$, and $B + D = -1$.
Adding the middle two equations gives $D + C = 1/2$, and so $B - A = 1/2$. Now $B + (C) = 1/2$ and so we obtain $2(B + C + D) = 0$, which leads to $B = -1/2$, $C = 1$, $D = -1/2$, and $A = -1$.
Hence the summand can be written as:
$$\frac{-r - 1/2}{r^2 + r + 1} + \frac{r-1/2}{r^2 - r + 1}$$
which is telescoping.
A: $$r^4+r^2+1=(r^2+1)^2-r^2=(r^2+r+1)(r^2-r+1)$$
Observe that if $f(n)=n^2+n+1, f(r-1)=(r-1)^2-(r-1)+1=r^2-r+1$
If $g(n)=an+b, g(n-1)=a(n-1)+b=an+b-a$
$$\text{Let  }\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(r)}{f(r)}-\dfrac{g(r-1)}{f(r-1)}$$ $$\text{so that the partial sum becomes }\sum_{r=1}^n\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(n)}{f(n)}-\dfrac{g(0)}{f(0)}$$
$$\text{Now, }\lim_{n\to\infty}\dfrac{g(n)}{f(n)}=\lim_{n\to\infty}\dfrac{an+b}{n^2+n+1}=\lim_{n\to\infty}\dfrac{\dfrac an+\dfrac b{n^2}}{1+\dfrac1n+\dfrac1{n^2}}=0\text{ for finite a,b}$$
$$\text{ and } \dfrac{g(0)}{f(0)}=?\text{  so we don't need the actual value of }a$$
$$\begin{align}\implies r^2-1=(ar+b)(r^2-r+1)-(ar+b-a)(r^2+r+1)=-ar^2+(a-2b)r+a\end{align}$$
$$\implies -a=1\iff a=-1, a-2b=0\iff b=\dfrac a2=-\dfrac12$$
