Thread: Derivative of an infinite series of functions

1. Derivative of an infinite series of functions

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

I know that I need to prove uniform convergence on (-1,1). so, Let $\displaystyle h_k(x)=\frac{(-1)^{n-1}}{n}$ and $\displaystyle g_k(x)=z^n$ for all $\displaystyle n\in\mathbb{N}$. I know that $\displaystyle h_k(z)$ converges to 0 and that $\displaystyle g_k(z)$ is always between (-1,1). So using the Dirichlet's test for uniform convergence i should be able to prove $\displaystyle f(z)$ is uniformly convergent and therefor $\displaystyle (\sum^\infty_{n=1}f_n(z))'=\sum^\infty_{n=1}f'_n(z )$. 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 $\displaystyle f_k,f : (-1,1) \to \mathbb{R}$ and $\displaystyle f_k(x_0)\to f(x_0)$ for one $\displaystyle x_0 \in (-1,1)$ and $\displaystyle f'\to g$ uniformly on compact subintervals of $\displaystyle (-1,1)$ then $\displaystyle f'=g$.

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