Differentiation under the integral sign and counting measure Consider a power series $f(x)=\sum_{n=1}^\infty a_nx^n$, and assume that $\displaystyle R=\lim_{n\to \infty}  \frac{a_n}{a_{n+1}}$ exists. 
Use differentiation under the integral sign to show that $f(x)$ is differentiable on $(-R,R)$ with derivative $f'(x)=\sum_{n=1}^\infty na_nx^{n-1}$. Hint: interpret this sum as an integral with respect to counting measure.
I am confused with this hint. How can I interpret this sum as an integral and what is counting measure?

The "sophisticated" solution:
$$f(x)=\sum_{n=1}^\infty a_nx^n$$
Here the sum is interpreted as the integral with respect to $\sum_{n=1}^\infty \delta_n$ of the function
$$
F(n,x) = a_n x^n
$$
Hence if we can apply the derivation theorem:
$$
f'(x) = \frac d{dx} \sum_{n=1}^\infty F(n,x) \color{red}= 
 \sum_{n=1}^\infty \frac d{dx}F(n,x) = \sum_{n=1}^\infty na_nx^{n-1}
$$
Now for the justification of the red '=' sign:
you have to check that, for example on $K=\left[-|x|,|x| \right]$ (on a compact subset of $(-R,R)$ containing $x$):
$$
\sum_{n=1}^\infty \sup_{x\in K} \left|\frac d{dx}F(n,x) \right| < \infty
$$
In this case, the sum is:
$$
=  \sum_{n=1}^\infty na_n |x|^{n-1}
$$
and this sum is finite as the radius of $
\sum na_nx^{n-1}
$
is also $R$.

One "elementary" solution:
Let $x\in (-R,R)$. The convergence of the series is uniform on $[0,x]$ (because $R$ is the radius of convergence of the series, and $[-x,x]$ is a compact subset of $(-R,R)$). Hence:
$$f(x) = \sum_{n=1}^\infty \int_0^x na_n t^{n-1} dt
= \int_0^x \sum_{n=1}^\infty  na_n t^{n-1} dt
$$
and then $f'(x) = \sum_{n=1}^\infty  na_n x^{n-1}$.
