Help me evaluate this series, please.

**smokesalot** · March 20th 2013, 12:22 PM

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

**MINOANMAN** · March 20th 2013, 12:37 PM

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

**smokesalot** · March 20th 2013, 12:50 PM

It does converge. It converges to ~ 1.9 but how can I show that properly?

**Soroban** · March 20th 2013, 03:23 PM

Hello, smokesalot!

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

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

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

$\text{We have: }\:\frac{(2r+1)(r^2+r+1)}{r(r+1)(r^4+r^2+1)}$

$\text{Multiply by }\frac{r-1}{r-1}\!:\;\;\frac{(2r+1)(r^2+r+1)}{r(r+1)(r^4+r^2+1)} \cdot\frac{r-1}{r-1} \;=\;\frac{(2r+1)(r^3-1)}{r(r^2-1)(r^4+r^2+1)}$

. . . . . . . $=\;\frac{(2r+1)(r^3-1)}{r(r^6-1)} \;=\;\frac{(2r+1)(r^3-1)}{r(r^3-1)(r^3+1)} \;=\;\frac{2r+1}{r(r^3+1)}$

Does that help?

**smokesalot** · March 21st 2013, 02:39 AM

Definitely helps, thank you!

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