$\sum_{n=0}^\infty n/2^{n}$ converges to show it? The infinite series $$\sum_{n=0}^\infty n/2^{n}$$ converges to $2$.
 But how to show that the infinite series converges to $2$?
Please help.
Thank you.
 A: Since $\frac{1}{2}<1$, you can interchange summation and differentiation in an infinite sum (uniform convergence). Also,  rewrite $ \frac{n}{2^n}$ as $p \frac{d}{dp}(p^n)$ where $p=\frac{1}{2}$. Can you handle from here?
EDIT You can hence rewrite the infinite sum as 
$$
p \frac{d}{dp}\sum_{k=0}^{\infty}p^{k}
$$
You get a simple geometric sum that you need to differentiate and multiply by $p$
A: Express $n/2^n$ as sum
$$\sum_{n=0}^\infty n/2^{n} = \sum_{n=1}^\infty \sum_{k=1}^n 2^{-n}$$
change order of sums
$$\sum_{k=1}^\infty \sum_{n=k}^\infty 2^{-n}$$
recall that $\sum_{n=k}^\infty 2^{-n} = 2^{1-k}$, and calculate
$$\sum_{k=1}^\infty 2^{1-k} = 2\sum_{k=1}^\infty 2^{-k} = 2$$
A: We know that for all $|x|<2$, 
$$\displaystyle \sum_{n=0}^{+\infty} \frac{x^n}{2^n} = \frac{1}{1-\frac{x}{2}}$$
Then, if you differentiate with respect to $x$ we get for all $|x|<2$
$$\displaystyle \sum_{n=0}^{+\infty} n\frac{x^{n-1}}{2^n} = \frac{1}{2(1-\frac{x}{2})^2}$$
so that
$$\displaystyle \sum_{n=0}^{+\infty} n\frac{x^{n}}{2^n} = \frac{x}{2(1-\frac{x}{2})^2}$$
and then do $x=1$.
