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?
 A: I guess you can use termwise differentiation here. Look at the derivative (wrt. $x$) of $\frac{1}{2^n-1+e^x}$. Then check if the series formed by using those expressions as terms is also uniformly convergent. If that is the case, the link seems to assure that $f$ is differentiable everywhere with that series as the derivative.
A: Given $f(x)=\sum_{n=0}^{\infty}\frac{1}{2^n-1+e^x}$, it follows that $f'(x)=\sum_{n=0}^{\infty}\frac{-e^x}{(2^n-1+e^x)^2}$.
\begin{align}
\left|\sum_{n=0}^{\infty}\frac{-e^x}{(2^n-1+e^x)^2}\right|\leq\\
C+\sum_{n=1}^{\infty}\left|\frac{e^x}{(2^n-1)^2}\right|=\\
C+e^x\sum_{n=1}^{\infty}\left|\frac{1}{4^n(1-2^{-n})^2}\right|<\infty\\
\end{align}
The conclusion follows from $1-2^{-n}\geq\frac{1}{2}$, for $n\geq1$.
We can't use the Weierstrass-M-test, because the estimation is dependent on $x$. But if we take a bounded subset of $\mathbb{R}$, we can estimate $e^x$. From this it follows that $f'(x)$ converges uniformly on every bounded subset of $\mathbb{R}$.
The last part: $f'(0)=\sum_{n=0}^{\infty}\frac{-1}{(2^n)^2}=-\frac{4}{3}$
A: 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
A: 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$$f'(0)=-\sum_{n=0}^{\infty}\frac{1}{4^n}=-\frac{4}{3}$$
