Q:

What is the sum of this by telescopic method?

$$\sum_{r=1}^\infty \frac{r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)}$$
This is to be done by telescopic method. I've used many things like writing $r^4+r^2+1=(r^2+r+1)(r^2-r+1)$ but have failed. I think all it needs is the correct rearranging of terms. 

A:

Hints:

Since $r^3+(r^2+1)^2=r(r^2+r)+(r^4+r^2+1)$,
$$
\frac{r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)}=\frac{r}{r^4+r^2+1}+\frac1{r^2+r}.
$$
Since $r^4+r^2+1=(r^2+r+1)(r^2-r+1)$,
$$
\frac{2r}{r^4+r^2+1}=a_r-a_{r+1},\qquad a_r=\frac{1}{r(r-1)+1}.
$$
Since $r^2+r=r(r+1)$,
$$
\frac1{r^2+r}=b_r-b_{r+1},\qquad b_r=\frac1r.
$$

And now, grab your telescope!

