Sum function of a power series derivative If $$f(x)=\sum_{n=1}^\infty \frac{2^n}{n} x^n\space\space\space x\in\left]-\rho,\rho\right[$$how is $$f'(x)=\frac{2}{1-2x}$$ I've found that $$\vert\rho\vert=\frac{1}{2}$$ and I know that$$\sum_{n=0}^\infty{x^n}=\frac{1}{1-x}\space\space\space x\in\left]-1,1\right[$$ (think I need to use that).
 A: lets say:
$$g(x)=\sum_{n=0}^\infty x^n$$
with the condition that $|x|<1$ in order for this to converge. Now look at:
$$g(2x)=\sum_{n=0}^\infty 2^nx^n$$
now think about what this sum evaluates to, and what the result would be from integrating
A: Letting $a_n=2^n/n$, we can differentiate term by term inside the radius of convergence, so we have
$$f'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}=\sum_{n=1}^{\infty}2^nx^{n-1}=2\sum_{n=0}^{\infty}(2x)^n=2\cdot\dfrac{1}{1-2x}$$
using the formula for the sum of a geometric series, which we can use since $|2x|<2\rho=1$.
A: Just try using the rules you know. the derivative is linear, which means you can just move it into finite sums (thanks @jjagmath), so we have to be careful. We have that
$$
\frac{\partial}{\partial x} f(x) =
\frac{\partial}{\partial x} \lim_{m \rightarrow \infty} 
\sum_{n=1}^{m} \frac{2^n}{n} x^n
$$
and want to compare it to
$$
\lim_{m \rightarrow \infty} \frac{\partial}{\partial x} \sum_{n=1}^{m} \frac{2^n}{n} x^n
=\lim_{m \rightarrow \infty} \sum_{n=1}^{m} \frac{2^n}{n} \frac{\partial}{\partial x} x^n
=\lim_{m \rightarrow \infty} \sum_{n=1}^{m} \frac{2^n}{n} n \ x^{n-1} \\
=\lim_{m \rightarrow \infty} \sum_{n=1}^{m} 2^n x^{n-1}
= \lim_{m \rightarrow \infty} 2 \ \sum_{n=1}^{m} 2^{n-1} x^{n-1}
= \lim_{m \rightarrow \infty} 2 \ \sum_{n=0}^{m-1} 2^{n} x^{n} \\
= \lim_{m \rightarrow \infty} 2 \ \frac{1-(2x)^m}{1-2x} 
= 2 \ \frac{1}{1-2x} =: g(x)
$$
the second to last equation is the finite geometric series which works if $|2x| < 1$ . Now all we need is an argument for why
$$
\frac{\partial}{\partial x} \lim_{m \rightarrow \infty} f_m(x) = \lim_{m \rightarrow \infty} \frac{\partial}{\partial x} f_m(x)
$$
But for this I'm not sure.
