Sum of $\sum_{n=1}^{\infty}\frac{n}{2^n}$. [duplicate]

I was trying to find the sum of $\sum_{n=1}^{\infty}\frac{n}{2^n}$.

I tried like this $S = \sum_{n=1}^{\infty}\frac{n}{2^n} = \sum_{n=1}^{\infty}(\frac{n+1}{2^n} - \frac{1}{2^n}) = 2\sum_{n=2}^{\infty}\frac{n}{2^n} - 1$

So,$S = 2(S - \frac{1}{2}) - 1$

Implying $S =2$.

EDIT:

Also from many posts in the comment below I found a method of looking at the sum $\sum x^n$ and differentiating and plugging for $x$.

Is there any other method of looking at the problem and calculating the sum apart from the above method,may be it would be interesting to see different approaches to the same problem which may be used to visualize other problems based on summation of series?

Multiply with $(2-1)$: $$(2-1)\sum_{n=1}^\infty n2^{-n}=\sum_{n=1}^\infty(n2^{1-n}-n2^{-n})=\sum_{n=0}^\infty (n+1)2^{-n}-\sum_{n=1}^\infty n2^{-n}=\sum_{n=0}^\infty2 ^{-n}$$

Summing $x^n$ gives $\frac{1}{1-x}-1$, so summing $nx^{n-1}$ gives $\frac{1}{(1-x)^2}$. Set $x=1/2$.

Hint: Use the geometric series

$$\frac{1}{1-u}=\sum_{n=0}^{\infty}u^n.$$

This series is uniformly convergent for $|u|\leq 1$, hence we can differentiate it with respect to $u$, in which $|u|\leq 1$.

$$-\frac{1}{(1-u)^2}=\sum_{n=1}^{\infty}n^{} u^{n-1} \implies -\frac{u}{(1-u)^2}=\sum_{n=1}^{\infty}n^{} u^{n} .$$

Relate this sum to your sum, in which $u=1/2$. Note, that your series is starting at $n=2$ and this series is starting at $n=1$.