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.
 A: 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 $$\sum_{k=1}^\infty k \left(\frac{1}{3}\right)^k=\sum_{k=0}^\infty k \left(\frac{1}{3}\right)^k-\sum_{k=0}^0 k \left(\frac{1}{3}\right)^k= \frac{\frac 13}{(1-\frac 13)^2}-0\,\cdot (\frac 13 )^0=\frac34$$
A: Let $S_n = \sum_{k = 1}^n k/3^k$. Then \begin{align}\frac{1}{3}S_n = \sum_{k = 1}^n \frac{k}{3^{k+1}} &= \sum_{k = 1}^n \frac{k+1}{3^{k+1}} - \sum_{k = 1}^n \frac{1}{3^{k+1}} = S_n - \frac{1}{3} + \frac{n+1}{3^{n+1}} + \frac{1}{3^2}\frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}}\\&= S_n - \frac{1}{3} + \frac{n+1}{3^{n+1}} +\frac{1}{6}\left[1 - \left(\frac{1}{3}\right)^n\right].\end{align} You can either isolate $S_n$, then take the limit as $n\to \infty$, or you can take the limit first to get an equation in $S_\infty := \lim_{n\to \infty} S_n$, then solve for $S_\infty$.
