Series $\sum_{n=0}^{\infty}\frac{1}{2^n-1+e^x}$ properties We need to prove $$f(x)=\sum_{n=0}^{\infty}\frac{1}{2^n-1+e^x}<\infty\ \ \forall_{x\in\mathbb{R}}$$
and then find all continuity points and all points in which $f$ is differentiable + calculate $f'(0)$. Especially the latter part is of difficult to me. It can be shown
$$\sum_{n=0}^{\infty}\frac{1}{2^n-1+e^x} <C+ \sum_{n=0}^{\infty}\frac{1}{2^n-3} < \infty$$ since $$\sum_{n=0}^{\infty}\frac{1}{2^n}=2$$ and $$\lim_{n\rightarrow\infty}\frac{\frac{1}{2^n-3}}{\frac{1}{2^n}} = \lim_{n\rightarrow\infty}\frac{2^n}{2^n-3} = \lim_{n\rightarrow\infty}\frac{1}{1-\frac{3}{2^n}} = 1.$$
Thus, using Weierstrass theorem, we conclude the initial sum is uniformly convergent for all $x\in\mathbb{R}$ to $f$, continuous on $\mathbb{R}$ (as a limit of uniformly convergent sequence of continuous functions). 
I hope it's more or less clear and correct. But how do I prove the second part of this exercise?

use ratio test,let $a_n=\frac{1}{2^n-1+e^x}$ and take,$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=0.5<1 $$so$\sum_{n=1}^{\infty}a_n$ converges
