$\displaystyle \sum_{r=1}^{\infty}\frac{r^3+(r+1)^3}{(r^4+r^2+1). (r^2+r)}=$

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- Mar 20th 2013, 12:22 PMsmokesalotHelp me evaluate this series, please.
$\displaystyle \sum_{r=1}^{\infty}\frac{r^3+(r+1)^3}{(r^4+r^2+1). (r^2+r)}=$

- Mar 20th 2013, 12:37 PMMINOANMANRe: Help me evaluate this series, please.
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 - Mar 20th 2013, 12:50 PMsmokesalotRe: Help me evaluate this series, please.
It does converge. It converges to ~ 1.9 but how can I show that properly?

- Mar 20th 2013, 03:23 PMSorobanRe: Help me evaluate this series, please.
Hello, smokesalot!

I don't have the sum yet, but I've done a**lot**of simplifying.

Quote:

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

$\displaystyle \text{Numerator:}$

. . . . . . . . $\displaystyle \overbrace{r^3 + (r+1)^3}^{\text{sum of cubes}} \:=\:\big[r + (r+1)\big]\,\big[r^2 - r(r+1) + (r+1)^2\big] \:=\:(2r+1)(r^2+r+1)$

$\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?

- Mar 21st 2013, 02:39 AMsmokesalotRe: Help me evaluate this series, please.
Definitely helps, thank you!