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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... –  J. M. Nov 15 '11 at 0:37
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 $\sum (e^{-x})^n$. –  André Nicolas Nov 15 '11 at 0:46
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? –  Arturo Magidin Nov 15 '11 at 1:19
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. –  M.Krov Nov 15 '11 at 3:56
@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 $x\gt 0$. –  Arturo Magidin Nov 15 '11 at 4:58

1 Answer 1

Since $$\,0<x\in\Bbb R \Longrightarrow e^{-x}=\frac{1}{e^x}<1$$ 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 $\,x>0\,$ .

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