Prove limit of $\sum_{n=1}^\infty n/(2^n)$ How do you prove the following limit?
$$\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{k}{2^k}\right)=2$$
Do you need any theorems to prove it?
 A: We may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by multiplying by $x$ and by making $n \to +\infty$ in $(2)$, using $|x|<1$, we get 
$$
\sum_{n=0}^\infty n x^n=\frac{x}{(1-x)^2}. \tag3
$$ Then put $x:=\dfrac12$.
Edit. One may observe we have avoided differentiating an infinite sum.
A: Hint:
$${\displaystyle {\frac {x}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n}\quad {\text{ for }}|x|<1\!}$$
A: 
I thought it might be useful to present two ways forward that rely on elementary pre-calculus knowledge only.  To that end, we proceed.

METHODOLOGY 1:
Note that we can write $n=\sum_{m=1}^n (1)$.  Therefore, 
$$\begin{align}
\sum_{n=1}^\infty nx^n&=\sum_{n=1}^\infty \sum_{m=1}^n(1) \,x^n\\\\
&=\sum_{m=1}^\infty \sum_{n=m}^\infty x^n\\\\
&=\sum_{m=1}^\infty \frac{x^m}{1-x}\\\\
&=\frac{x}{(1-x)^2}
\end{align}$$

METHODOLOGY 2:
Let $S=\sum_{n=1}^\infty nx^n$. Note that we can write 
$$\begin{align} x S&=\sum_{n=1}^\infty nx^{n+1}\\\\
&=\color{blue}{\sum_{n=1}^\infty (n+1)x^{n+1}}-\color{red}{\sum_{n=1}^\infty x^{n+1}}\\\\
&=\color{blue}{S-x}-\color{red}{\frac{x^2}{1-x}}\\\\
(1-x)S&=x+\frac{x^2}{1-x}\\\\
S&=\frac{x}{(1-x)^2}
\end{align}$$
as expected!
A: The series is obviously convergent. Set $S:=\sum_{n=1}^\infty \frac{n}{2^n}$. This gives
$$S=\sum_{n=1}^\infty \frac{n}{2^n}= \sum_{n=1}^\infty \frac{n-1}{2^{n-1}}+\sum_{n=1}^\infty \frac{1}{2^{n-1}} =\frac{1}{2} \sum_{n=1}^{\infty}\frac{n}{2^n}+ \sum_{n=1}^{\infty}\frac{1}{2^n}=\frac{S}{2} + \frac{1}{2}\ \frac{1}{1-\frac{1}{2}}=\frac{S}{2}+1$$
and therefore
$$S=\frac{S}{2}+1$$
that is $S=2$.
