Calculating the sum of $f(x) = \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^n}$ In Calculus, how do I calculate this sum?
$$f(x) = \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^n}$$
This is what I did so far:
$$ f(x) = 2x \cdot \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^{n-1}} $$
Therefore:
$$ \frac{\int{f(x)}}{2x} = \sum_{n=0}^{\infty} {2^n \cdot x^n}$$
But I have no idea where to continue from here!
 A: I believe you can sum the series at the point you left off,
$$
\frac{\int f(x) dx}{x} = 1 + 2x + (2x)^2 + (2x)^3 + ... = \frac{1}{1-2x}
$$
So, we work back to $f$ by itself,
$$
\int f(x) dx = \frac{x}{1-2x}
$$
We take derivatives,
$$
f(x) = \frac{(1-2x) - x (-2)}{(1-2x)^2} = \frac{1}{(1-2x)^2}
$$
A: $$\sum_{n=1}^\infty nt^n=t\frac{d}{dt}\left(\sum_{n=1}^\infty t^n\right)=t\frac{d}{dt}\left(\frac{t}{1-t}\right)=\frac{t}{(1-t)^2}$$
Now with $t=2x$ we find
$$\sum_{n=1}^\infty n(2x)^n=\frac{2x}{(1-2x)^2}$$
A: Here is a more elementary proof, which doesn't use the differentiability/integrability property of power series:
Let 
$$S_m= \sum_{n=0}^{m} {n \cdot 2^n \cdot x^n}$$
Then 
$$2x S_m =\sum_{n=0}^{m} {n \cdot 2^{n+1} \cdot x^{n+1}}=\sum_{k=1}^{m+1} {(k-1) \cdot 2^{k} \cdot x^{k}}$$
$$=\left(\sum_{k=1}^{m+1} k \cdot 2^{k} \cdot x^{k}\right)-\left(\sum_{k=1}^{m+1}   2^{k} \cdot x^{k}\right)=\left(\sum_{k=0}^{m+1} k \cdot 2^{k} \cdot x^{k}\right)-\left(2x\cdot\frac{1-(2x)^{m+1}}{1-2x}\right)$$
$$2xS_m=\left(S_m+(m+1)2^{m+1}x^{m+1}\right)-\left(2x\cdot\frac{1-(2x)^{m+1}}{1-2x}\right)$$
Solving for $S_m$ yields:
$$S_m=\left(2x\cdot\frac{1-(2x)^{m+1}}{(1-2x)^2}\right)-\frac{(m+1)(2x)^{m+1}}{1-2x}$$
now, $S_m$ is convergent if and only if $(2x)^{m+1} \to 0$ if and only if $|2x|<1$.
In this case
$$\lim S_m= 2x\cdot\frac{1}{(1-2x)^2}$$
