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

  • 2
    $\begingroup$ I think this question and extremely similar ones have been asked too many times, time to use the search feature. $\endgroup$ Commented Sep 12, 2015 at 17:03

1 Answer 1


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 . $\endgroup$
    – Asinomás
    Commented Sep 12, 2015 at 16:58

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