# Thread: Derivative of an infinite series of functions

1. ## Derivative of an infinite series of functions

Let $f(z)=\sum^\infty_{n=1}\frac{(-1)^{n-1}z^n}{n}$, $-1. Prove that $f'(z)=\sum^\infty_{n=1}(-1)^{n-1}z^{n-1}$ on (-1,1).

I know that I need to prove uniform convergence on (-1,1). so, Let $h_k(x)=\frac{(-1)^{n-1}}{n}$ and $g_k(x)=z^n$ for all $n\in\mathbb{N}$. I know that $h_k(z)$ converges to 0 and that $g_k(z)$ is always between (-1,1). So using the Dirichlet's test for uniform convergence i should be able to prove $f(z)$ is uniformly convergent and therefor $(\sum^\infty_{n=1}f_n(z))'=\sum^\infty_{n=1}f'_n(z )$. Will this work? and how do i write a formal proof if it is?

2. ## Re: Derivative of an infinite series of functions

Uniform convergence of a sequence doesn't guarantee even pointwise convergence of the sequence of derivatives, on the other hand you have the following:

If $f_k,f : (-1,1) \to \mathbb{R}$ and $f_k(x_0)\to f(x_0)$ for one $x_0 \in (-1,1)$ and $f'\to g$ uniformly on compact subintervals of $(-1,1)$ then $f'=g$.

In your case take $x_0=0$ and since the sequence (series) of derivatives is a geometric series the result is immediate.