Finding of $\sum_{n=0}^{\infty} n\left(\frac12\right)^n$ Evaluate $$\sum_{n=0}^{\infty} n\left(\frac12\right)^n$$
 A: Here is a useful finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ you get
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ and by making $n \to +\infty$ in $(1)$ and $(2)$, using $|x|<1$, gives 
$$
1+x+x^2+...+x^n+...=\frac{1}{1-x} \tag3
$$
$$
1+2x+3x^2+...+nx^{n-1}+...=\frac{1}{(1-x)^2} \tag4
$$
If you set $x=\frac12$ in $(4)$, you obtain an answer to your question.
A: Let $S = \sum_{n = 0}^\infty n(1/2)^n$. Then $$\frac{1}{2}S = \sum_{n = 0}^\infty n\left(\frac{1}{2}\right)^{n+1} =  \sum_{n = 0}^\infty (n + 1)\left(\frac{1}{2}\right)^{n+1} - \sum_{n = 0}^\infty \left(\frac{1}{2}\right)^{n+1} = S - \frac{1/2}{1 - 1/2} = S - 1$$
Solving for $S$ yields $S = 2$.
A: Imagine your series as a 2-dimensional series of series that you are adding up, where the first series (row) is $\sum_{n=1}^\infty (1/2)^n$ and the second series (row) is $\sum_{n=2}^\infty (1/2)^n$ the third series (row) is $\sum_{n=3}^\infty (1/2)^n$, and so on forever. Clearly this collection of series rows equals your original series if you add up equal powers $(1/2)^n$ across rows. But now note that you can factor out a power of $(1/2)^k$ from the $k$th row, and when you do you will always get $\sum_{n=0}^\infty (1/2)^n$ remaining, which you  should know the formula for. So then your overall answer is $(1/2 + 1/2^2 + 1/2^3 + \ldots) = \sum_{n=1}^\infty (1/2)^n$ which you should know a formula for, multiplied by that infinite sum formula for $\sum_{n=0}^\infty (1/2)^n$.
A: Hint: $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ for $|x| < 1$ 
Now differentiate both sides and substitute $x=0.5$, what do you get? 
A: $$1+x+x^2+...+x^n=\frac{1}{1-x}$$
differentiate both sides to get
$$1+2x+3x^2+4x^3+....=\frac{1}{(1-x)^2}$$
multiply both sides by $x$
$$x+2x^2+3x^3+4x^3+....=\frac{x}{(1-x)^2}$$
