Find the Sum of an Infinite Series Compute
$${1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.}$$
I tried creating a partition for this but no such luck.
What would the the equation to generate the sum for the nth term
 A: One may recall 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
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ multiplying by $x$ and making $n \to +\infty$ in $(2)$, using $|x|<1$, gives 
$$
\sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}. \tag3
$$ Then put $x=\dfrac12$ in $(3)$.
A: Let
$$f(x)=\sum_{n=0}^{+\infty}\frac{x^n}{2^n}=\frac{2}{2-x}$$
with a radius of convergence $R=2$
$$f'(x)=\sum_{n=0}^{+\infty}n\frac{x^{n-1}}{2^n}=\frac{2}{(2-x)^2}$$
thus, your sum is $f'(1)=2$.
A: Let $S=\sum_{n\geq 1}\frac{n}{2^n}.$ Then 
$$2S=\sum_{n\geq 1}\frac{2n}{2^n}=\sum_{n\geq 1}\frac{n}{2^{n-1}}=\sum_{n\geq 1}\frac{(n-1)+1}{2^{n-1}} = 2+\sum_{n\geq 1}\frac{n-1}{2^{n-1}} = 2+S $$
hence $S=2$.
A: 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/(2^n) + ..... = (1/2)*2 = 1
  1/4 + 1/8 + 1/16 + ..... + 1/(2^n) + ..... = [1/(2^2)]*2 = 1/2

        1/8 + 1/16 + ..... + 1/(2^n) + ..... = [1/(2^3)]*2 = 1/(2^2)

So on and so forth.
Hence, S = 1 + 1/2 + 1/(2^2) + ..... + 1/(2^n) + ..... = 1 + 1 = 2
