# Double differentiable

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..?

• 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. – Ethan Bolker Jun 3 '19 at 14:13
• @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? – mahma Jun 3 '19 at 14:23
• 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? – Ethan Bolker Jun 3 '19 at 14:25
• @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 $|x|<r$? – mahma Jun 3 '19 at 14:46
• Yes. Now you can post an answer here to your own question, so that it does not remain on the queue attracting attention. – Ethan Bolker Jun 3 '19 at 15:34

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