Using the formula $\sum^{\infty}_{n=0} {x^n} = \frac{1}{1-x}$ take the derivative of both sides with respect to the variable x to find a new formula for another series. Use this formula to compute exactly the series:
$\sum^{\infty}_{n=0} \frac{n}{3^n} = \frac{1}{3}+\frac{2}{9} + \frac{3}{27} + \frac{4}{81}$
What is this question asking me to do? I don't know of a way to take the derivative of $\sum^{\infty}_{n=0} {x^n}$, but that is probably just my lack of experience. I am sorry for the lack of attempt to answer this question, I am completely lost.
Note that term by term differentiation followed by multiplication by $x$ gives us that $$ \frac{x}{(1-x)^2}=xD\left(\frac{1}{1-x}\right)=x\sum_{n=0}^\infty nx^{n-1}=\sum_{n=0}^\infty nx^{n} \quad (|x|<1). $$