How to deal with subtraction of sigma? How to solve this?

$$\displaystyle\sum_{n=1}^\infty
 \frac{n}{3^{n-1}}-\displaystyle\sum_{n=1}^\infty \frac{n}{3^n}$$

My work
$$S=\displaystyle\sum_{n=1}^\infty \left(\frac{n}{3^{n-1}}-\frac{n}{3^n}\right)$$
$$S=\displaystyle\sum_{n=1}^\infty\frac{3n-n}{3^n}=\displaystyle\sum_{n=1}^\infty \frac{2n}{3^n}$$
which seems to yield nothing.
 A: I suspect that you're supposed to shift the index of one of the series. What you've done is 100% valid, but as you said, it's not yielding something "useful". Instead, take the series
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}},$$
and shift the index. Let $m = n - 1$. Then when $n = 1$, $m = 0$, so
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}} = \sum_{m=0}^\infty \frac{m+1}{3^{m}} = \sum_{n=0}^\infty \frac{n + 1}{3^n} = 1 + \sum_{n=1}^\infty \frac{n + 1}{3^n}.$$
Thus,
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}}-\sum_{n=1}^\infty \frac{n}{3^n} = 1 + \sum_{n=1}^\infty \frac{n + 1}{3^n} - \sum_{n=1}^\infty \frac{n}{3^n} = 1 + \sum_{n=1}^\infty \frac{1}{3^n},$$
which is a geometric series that you can calculate easily.
A: I consider the other answers as superior to the approach that I am going to document.  I offer it merely as an alternative perspective.
For $r \in (0,1),~~$ let $~~A = \sum_{k = 0}^\infty r^k = \frac{1}{1-r}~~$ and let $~~B = \sum_{k = 1}^\infty r^k = \frac{r}{1-r}.$
Let $T = \sum_{k=1}^\infty kr^k ~~\implies 
~~T = r + 2r^2 + 3r^3 + 4r^4\cdots $
$= B + rB + r^2B + r^3B + \cdots = B(1 + r + r^2 + r^3 + \cdots) = BA.$
A: You can do differentiation on series. So if you know (this is a basic geometric series)
$$\sum_{n=1}^{\infty}r^n = \frac{r}{1-r}$$
Then differentiate with respect to $r$.
$$\sum_{n=1}^{\infty}nr^{n-1} = \frac{1}{(1-r)^2}$$
or, by multiplying by $r$,
$$\sum_{n=1}^{\infty}nr^{n} = \frac{r}{(1-r)^2}$$
Do you see how your series is reflected in this formula?
