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Problem:

Prove that the series $f\left ( x \right )= \sum_{n=0}^{\infty } e^{-n x}$ converges for all $x>0$ and that $f$ is infinitely differentiable on $(0, \infty)$.

My solution is: I treated the sum as a sum of geometric series $f\left ( x \right )= \sum_{n=0}^{\infty } \left ( e^{-x } \right )^{n}$ and then $f(x)= \frac{1}{1- e^{-x}}$. Does this prove the convergence of f for all $x\in \mathbb{R}$?

Now I am stuck on how to prove that $f$ is infinitely differentiable. Can you guys give me a detailed proof? Thanks

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    May I suggest writing an answer to your onw question if you genuinely understood our comments?2011-11-15

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Since $\,0 we get a convergent infinite geometric series $\sum_{n=0}^\infty e^{-nx}=\sum_{n=0}^\infty \left(\frac{1}{e^x}\right)^n=\frac{1}{1-e^x}$

Thus $f(x)=\sum_{n=0}^\infty e^{-nx}=\frac{1}{1-e^x}\,\,,\,x>0$ is infinitely differentiable (= i.d.) since it is the composition $f(x)=g\circ h(x)\,\,,\,\,h(x):=e^x\,\,,\,g(x):=\frac{1}{1-x}$ and both functions $\,g,h\,$ are i.d. for $\,x>0\,$ .