Calculate the sum of $\sum^{\infty}_{n=0} (-1)^n \frac{n+1}{3^n}$. Currently going through old Analysis I - exams and I'm having a problem witht his one:

Calculate the sum of the series: $$\sum^{\infty}_{n=0} (-1)^n \frac{n+1}{3^n}$$

What I did was firstly to rearrange the sequence:
$$\sum^{\infty}_{n=0} (-1)^n \frac{n+1}{3^n} = \sum^{\infty}_{n=0} (n+1) \left(-\frac{1}{3}\right)^n $$
But with $(n+1)$ I cannot apply the formula for a geometric series, and I don't know how to continue from here.
 A: For $\left| x\right|\lt 1$ you can write
$$\sum _{n=0}^{\infty}x^{n+1} = x \sum_{n=0}^{\infty} x^n = \frac{x}{1-x}\tag{1}$$
Now it must hold
$$\sum_{n=0}^{\infty} \left(n+1\right)x^n =\frac{d}{dx}\sum_{n=0}^{\infty} x^{n+1} \stackrel{(1)}{=}\frac{d}{dx} \frac{x}{1-x} = \frac{1}{(1-x)^2}$$
Thus for $x=-\frac{1}{3}$ 
you will get
$$\sum_{n=0}^{\infty} \left(n+1\right)\left(-\frac{1}{3}\right)^n = \frac{1}{\left(1+\frac{1}{3}\right)^2} = \frac{9}{16}$$
A: Notice that
$$
\sum_{n=0}^\infty(-1)^n\frac{n+1}{3^n}=f\left(-\frac13\right),
$$
with
$$
f(x)=\sum_{n=0}^\infty (n+1)x^n.
$$
For every $x\in (-1,1)$ we have
$$
f(x)=\sum_{n=0}^\infty (n+1)x^n=
\frac{d}{dx}\left[x\sum_{n=0}^\infty x^n\right]=\frac{d}{dx}\left(\frac{x}{1-x}\right)=\frac{1}{(1-x)^2} \quad \forall x\in (-1,1).
$$
Therefore
$$
\sum_{n=0}^\infty(-1)^n\frac{n+1}{3^n}=\frac{1}{(1+1/3)^2}=\frac{9}{16}.
$$
A: The general method to attack this kind of problem is to use power series, as in the other answers. But for this particular problem, no such machinery is necessary, as hinted by JLamprong in the comments. 
Indeed, calculating $S+S/3$ one gets
$$
S+\frac{S}3=\sum_{n=0}^\infty(-1)^n\frac{n+1}{3^n}+\sum_{n=0}^\infty(-1)^n\frac{n+1}{3^{n+1}}\\
=\sum_{n=0}^\infty(-1)^n\frac{n+1}{3^n}-\sum_{n=1}^\infty(-1)^n\frac{n}{3^{n}}\\
=\sum_{n=0}^\infty(-1)^n\frac{1}{3^n}=\frac1{1+\frac13}=\frac34.\\
$$
So $4S/3=3/4$, i.e. $S=9/16$.
