$\displaystyle \sum_{r=1}^{\infty}\frac{r^3+(r+1)^3}{(r^4+r^2+1). (r^2+r)}=$
Smokesalot
the general term tends to zero as n goes to infinity this means that the series might converge.
try to use one of the known tests and verify if it converges or not.
check here for convergence tests Convergence tests - Wikipedia, the free encyclopedia
MINOAS
Hello, smokesalot!
I don't have the sum yet, but I've done a lot of simplifying.
$\displaystyle \displaystyle\sum_{r=1}^{\infty}\frac{r^3+(r+1)^3} {(r^4+r^2+1)(r^2+r)}$
$\displaystyle \text{Numerator:}$
. . . . . . . . $\displaystyle \overbrace{r^3 + (r+1)^3}^{\text{sum of cubes}} \:=\:\big[r + (r+1)\big]\,\big[r^2 - r(r+1) + (r+1)^2\big] \:=\:(2r+1)(r^2+r+1)$
$\displaystyle \text{We have: }\:\frac{(2r+1)(r^2+r+1)}{r(r+1)(r^4+r^2+1)} $
$\displaystyle \text{Multiply by }\frac{r-1}{r-1}\!:\;\;\frac{(2r+1)(r^2+r+1)}{r(r+1)(r^4+r^2+1)} \cdot\frac{r-1}{r-1} \;=\;\frac{(2r+1)(r^3-1)}{r(r^2-1)(r^4+r^2+1)} $
. . . . . . . $\displaystyle =\;\frac{(2r+1)(r^3-1)}{r(r^6-1)} \;=\;\frac{(2r+1)(r^3-1)}{r(r^3-1)(r^3+1)} \;=\;\frac{2r+1}{r(r^3+1)} $
Does that help?