# Proving an infinite series is differentiable

I'm trying to prove that the following series is differentiable with respect to $x$ on any interval $[a,b]$ with $0<a<b$:

$$f(x)=\sum_{n=1}^\infty \frac{e^{-nx}}{n}$$

What I've tried is applying the Term-wise Differentiability Theorem, where I need to show that the following three conditions hold:

1. Each $f_n$ is differentiable on $[a,b]$
2. $\sum_{i=1}^n f_i'$ converges to $g$ uniformly on $[a,b]$ (This is where I really get stuck)
3. $\sum_{i=1}^n f_i(x_0)$ converges for some $x_0$ on [a,b]

I already get stuck on choosing $f_n$ and going on with step 1 and 2, and I don't know of any other way to show that the infinite series is differentiable with my current knowledge, so I hope someone could help me out.

• I assume that sum is over $n$ from $1$ to $\infty$, not over $i$ from $1$ to $n$? – Jack M Jun 6 '17 at 14:55
• Yes that is definitely true, this was my first attempt at using Mathjax. My mistake, thanks. – markovmontecarlo Jun 6 '17 at 21:13

## 1 Answer

$$f'(x)=\sum_n \frac{d}{dx}\frac{e^{-nx}}n=-\sum_n e^{-nx}=-\sum_n (e^{-x})^n$$

Does that give you any ideas?

(Note: in order to be rigorous, the first step of distributing the derivative over the sum should be done for a partial sum, since we don't know yet if the sum of derivatives converges)

• So simple, yet it helped me a ton. I was probably quite tired, as I did not see this immediately. Thank you very much! – markovmontecarlo Jun 6 '17 at 21:15