I have a series from n=1 to infinity of
n/(4^n).
Can this convert into a geometric series? and so how?
Yes it can by differentiation.
Start with $\displaystyle \displaystyle f(x) = \sum\limits_{n = 0}^\infty {x^n } = \frac{1}{{1 - x}},~|x|<1$.
Differentiate to show $\displaystyle \displaystyle f'(x) = \sum\limits_{n = 1}^\infty {nx^n } = \frac{x}{{(1 - x)^2}}$
Then let $\displaystyle x=\frac{1}{4}$
You let $\displaystyle x = \frac{1}{4}$ and the LHS, that's the sum $\displaystyle \sum\limits_{n = 1}^\infty {nx^n }$ becomes $\displaystyle \sum\limits_{n = 1}^\infty {\frac{n}{4^n} }$ which is your original sum.
Do the same, i.e. let $\displaystyle x = \frac{1}{4}$ for the RHS, that's $\displaystyle \dfrac{x}{{(1 - x)^2}}$ and you will find what it evaluates to.