How to find the value this sum converges to?$\sum_{n=0}^{\infty}n·\frac{3^{n-1}}{4^{3n+1}} $

Question

How to find the value this sum converges to?$$\sum_{n=0}^{\infty}n·\frac{3^{n-1}}{4^{3n+1}} $$ I've no clue how to simplify $ \frac{3^{n-1}}{4^{3n+1}} $ so if you could help me with that I could do the rest by myself so any hint would be appreciated.

Answer 1

notice that

$$ \frac{ 3^{n-1} }{4^{3n+1} } = \frac{1}{12} \left( \frac{3}{64} \right)^n $$

Now, use the fact that

$$ \sum n x^n = \frac{x}{(1-x)^2} , \; \; \;\ |x|<1$$

as one can easily see that upon differentiating the geometric sum $\sum x^n = \frac{1}{1-x} $ with respect to $x$ and then multiplying by $x$ and since the radius of convergence remains the same upon differentition and the result follows