infinite series sum (1 Viewer)

Lieutenant_21 · Mar 21, 2013

1.9478660887918053629526007244099960758086061625067056 (don't ask for steps :$)

Sy123 · Mar 21, 2013

If it helps other people, I have been able to reduce the summand to:

$\frac{1}{r^2-r+1} (\frac{1}{r} + \frac{1}{r+1})$ and unfortunately I can't see a telescoping sum here :/

I'm still working on the problem though

Lieutenant_21 · Mar 21, 2013

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

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

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

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

$continuing...$

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infinite series sum (1 Viewer)

juantheron

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Lieutenant_21

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Sy123

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Lieutenant_21

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