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

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.

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