Limit of $\sum_{k=1}^{n} \frac{k}{3^k}$ I need to calculate $$\lim_{n \to \infty}\sum_{k=1}^{n} \frac{k}{3^k}$$ and I can't really do it. I have a feeling it's simple and that there's a simple catch but I just can't see it. If anyone could provide me with a hint to solve this, I would be grateful.
I know this question is classified as homework and that it seems like I haven't tried anything, but I have. I tried expanding this but that doesn't help at all. Geometric progression came to mind at first, but I realized it's not very useful either. 
Thanks.

Knowing that $$\sum_{k=0}^\infty n^k=\frac{1}{1-n}$$
$$\implies \frac{d}{dn}\sum_{k=0}^\infty n^k=\frac{d}{dn} \frac{1}{1-n}$$
$$\implies \sum_{k=0}^\infty kn^{k-1}=\frac{1}{(1-n)^2}$$
$$\implies \sum_{k=0}^\infty kn^k=\frac{n}{(1-n)^2}$$
Your sum is $$\sum_{k=1}^\infty k \left(\frac{1}{3}\right)^k=\sum_{k=0}^\infty k \left(\frac{1}{3}\right)^k-\sum_{k=0}^0 k \left(\frac{1}{3}\right)^k= \frac{\frac 13}{(1-\frac 13)^2}-0\,\cdot (\frac 13 )^0=\frac34$$
