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I wanted to compute the radius of convergence for the following the power series

$$\sum_{n=1}^{\infty} a_nz^n$$

with $(i) \, a_n = n!, \, (ii) \, a_n = \sqrt[\leftroot{-3}\uproot{3}n]{n}$

Then I need to determine which power series defines a function that is twice differentiable and compute the second derivative.

For the convergence radius I had no problem, and used once the Alembert criterion, once the Cauchy criterion and found that the convergence radius for $a_n = n!$ is 0, and the one for $a_n = 1$.

Now I'm tempted to say that therefore the first power series gives a non differentiable function, since it doesn't converge. Is that right? Because I'm not sure about it.

Then for the second I was told that since the convergence radius is $1$, it's differentiable at $]-1,1[$, which I can understand, but that it's also automatically differentiable twice at $]-1,1[$. And that part I don't undertand. Why? Has that something to do with the radius being specifically $1$? And how to conclude in general?

For the last question, about derivating twice, isn't it just derivate the sum? So I'd get:

$$f''(x) = \sum_{n = 2}^{\infty} n(n-1)a_nz^{n-2}$$

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Every convergent power series is infinitely differentiable in the same interval of convergence. $$ f'(x)=\sum_{n=1}^{\infty}na_nx^{n-1} $$ So radius of convergence of $f'(x)$ is $$ {1\over\limsup(n|a_n|)^{1\over n}}={1\over\limsup |a_n|^{1\over n}} $$ as $$\lim_{n\to\infty}n^{1\over n}=1$$

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  • $\begingroup$ Ok, so "at least twice differentiable" meant exactly that. Is there a way that the power series is exactly once differentiable? $\endgroup$
    – K.A.
    Jan 28, 2016 at 22:48
  • $\begingroup$ nope ........:) $\endgroup$ Jan 28, 2016 at 22:50
  • $\begingroup$ Ok. Then it was a weird question. Thank you anyway :). $\endgroup$
    – K.A.
    Jan 28, 2016 at 22:55

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