Several part sum of power series question

(a) Prove that $\sum_{j=0}^{\infty}x^j$ is differentiable on $(-1,1)$ and $\frac {d}{dx}\sum_{j=0}^{\infty}x^j = \sum_{j=0}^{\infty}(j+1)x^j$.

(b) Use the fact that $\sum_{j=0}^{\infty}x^j = \frac {1}{1-x}$ on $(-1,1)$ to find a formula for $\sum_{j=0}^{\infty}(j+1)x^j$

(c) Use this to calculate $\sum_{j=0}^{\infty}\frac {j+1}{2^j}$ exactly.

work for part (a)

We can represent $\sum_{j=0}^{\infty}x^j$ with an integral, therefore it is differentiable?

Do I just need to prove that $jx^{j-1}=(j+1)(x^j)$?

In which case $j=\frac{x^2}{1-x}$..?

(b) Using $j=\frac{x^2}{1-x}$

and maybe using $\frac{d}{dx}\frac{1}{1-x} = \frac{1}{(1-x)^2}$..

Not sure how to put that all together though :(

(c) Do I use $j=\frac{x^2}{1-x}$ and then just plug and chug?

I'm really not sure how I'm supposed to go about doing this as you can tell.. :/

a) The radius of convergence of the given series is $R=1$ . This can be computed by Cauchy-Hadamard formula.
Then $f(x)=\sum_{j=0}^{\infty}x^j$ is continuous on $(-1,1)$, and differentiable on $(-1,1)$ except possibly at its endpoints.
c) Subsitute $x=\frac{1}{2}$