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Given$$\sum_{n=2}^\infty \frac{x^n}{n^2-n}$$ Explain why the series sum function $f$ is two times differentiable in the interval (-1,1) with $$f''(x)=\frac{1}{1-x}$$

I have already showed that the second derivative is equal $\frac{1}{1-x}$, but let's say I need to explain why it is differentiable. Is it correct to say that the function is two times differentiable because n starts from 2 and therefore will have $x^2$ in the numerator, which will derive respectively to $2x, 2, 0$ and still be differentiable for all $x \in (-1,1)$? Am I on the right track or..?

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    $\begingroup$ There is a general theorem that says you can differentiate a power series term by term inside its interval of convergence. If that theorem is available to you all you have to do is find the radius of convergence. $\endgroup$ – Ethan Bolker Jun 3 '19 at 14:13
  • $\begingroup$ @EthanBolker I found that the radius of convergence is 1. My curriculum book only shows that it is differentiable for the first derivative for $|x| < r = 1$, but says nothing about the second, but I assume it is also applicable for the second derivative? $\endgroup$ – mahma Jun 3 '19 at 14:23
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    $\begingroup$ What do you know about the radius of convergence of the differentiated series? What does that tell you about how many times you can differentiate? $\endgroup$ – Ethan Bolker Jun 3 '19 at 14:25
  • $\begingroup$ @EthanBolker So the first derivative will have the same radius of convergence as the original power series and due to the fact that the $r>0$ the sum of power series will actually be infinitely differentiable for all $|x|<r$? $\endgroup$ – mahma Jun 3 '19 at 14:46
  • $\begingroup$ Yes. Now you can post an answer here to your own question, so that it does not remain on the queue attracting attention. $\endgroup$ – Ethan Bolker Jun 3 '19 at 15:34
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Find the radius of convergence which will tell you whether or not the series sum function is differentiable.

Without going into details, every power series has a radius of convergence $0 \leq R \leq \infty$. The power series will converge for $|x|<R$ and diverge for $|x|>R$. A theorem says that in case of $R>0$, the sum of the power series will be infinitely differentiable in $|x|<R$. Bare in mind that the radius of convergence remains unchanged as you derive the series sum function.

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