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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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    If you can show why the quantity $e^{-x}$ is within the domain of convergence of the geometric series, then your proof is complete. For infinitely differentiable: you've composed two functions that are themselves infinitely differentiable; thus...2011-11-15
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    Not for all $x$ in $\mathbb{R}$, the question only asked you to do it for $x>0$, and in fact the series diverges for $x \le 0$. The expression $\sum (e^{-n})^x$ is technically correct but you really want $\sum (e^{-x})^n$.2011-11-15
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    If you want to treat it as a geometric series, shouldn't you write it as $f(x) = \sum_{n=0}^{\infty}(e^{-x})^n$ instead?2011-11-15
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    Thanks all for your comments. yes, I meant to write it as $f\left ( x \right )= \sum_{n=0}^{\infty } \left ( e^{-x } \right )^{n}$. So, for convergence I know how to do it. How about infinitely deffrentiable? J.M says f(x) is a composition of two infinitely differentiable functions. Can anyone tell what these two functions are? OR if anyone of you has a different way to show this, please let me know.2011-11-15
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    @Zi2018Alpha: One function is $x\mapsto e^{-x}$; the other is $y\mapsto \frac{1}{1-y}$, which is infinitely differentiable on $(0,1)$, which is where $e^{-x}$ is when $x\gt 0$.2011-11-15
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    Yes, Indeed with those two infinitely differentiable functions, we can form a function f(x) which is a composition of the previous two functions and hence it is infinitely differentiable function. Thanks a lot for your help.2011-11-15
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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

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 $\,x>0\,$ .