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The question is:

Find the value of $ \sum_{n=1}^{\infty} \frac{n^3}{3^n} $. I know this sum converges and that it's value is $ \frac{33}{8} $, however, I can't seem to find it.

I've tried doing $3S-S$ to try and find a pattern, tried using different subtractions, all to no avail.

Any help would be gladly accepted.

Thanks in advance, Pedro


marked as duplicate by Jack D'Aurizio, user147263, Community Sep 12 '15 at 20:55

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    $\begingroup$ I think this question and extremely similar ones have been asked too many times, time to use the search feature. $\endgroup$ – Jack D'Aurizio Sep 12 '15 at 17:03

Hint: $$ \sum_0^∞ x^n = \frac{1}{1-x} $$ implies by differentiation wrt $x$: $$\sum_1^∞ nx^{n-1} = \frac{\text{d}}{\text{d}x}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2} $$

  • $\begingroup$ good hint. OP, please tell me if you have trouble finishing from here . $\endgroup$ – Jorge Fernández Hidalgo Sep 12 '15 at 16:58

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