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?

well let, $$g_n(x)=\frac{1}{2^n-1+e^x}$$for n={0,1,2,3..}
;we can check the differentiability of this function easily and it is indeed differentiable.so$$f(x)=\sum_{n=0}^{\infty}g_n(x)$$so,obviously f(x) is also differentiable and hence continuous at all real numbers$$f'(0)=-\sum_{n=0}^{\infty}\frac{1}{4^n}=-\frac{4}{3}$$
