Find the value of $ \sum_{n=1}^{\infty} \frac{n^3}{3^n} $

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friends.

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

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

score 5 · Answer 1 · answered Sep 12 '15 at 16:54

5

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

answered Sep 12 '15 at 16:54

Calvin Khor

34,903

good hint. OP, please tell me if you have trouble finishing from here . – Asinomás Sep 12 '15 at 16:58

Find the value of $ \sum_{n=1}^{\infty} \frac{n^3}{3^n} $

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