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.

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.
