sequences and series - 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?

Answer

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.