1
$\begingroup$

I have a question.

I want to know an example of a power series centered at $x = 0$ that converges on $[-2,2]$ but not absolutely on the entire interval $[-2,2]$, and diverges otherwise.

I saw that it could be $\frac{(-1)^k}{k}x^{k(k+1)}$. But I don't know how you can say that it converge on $[-2,2]$ because with the ratio test or the root test I get only an interval of $(-1,1)$? And do they mean that it is not absolutely convergent on the entire interval $[-2,2]$ that it converge conditionally?

Can someone help me? Thank you

  • 0
    Also, you need to be careful with the ratio test when your series is missing terms. It doesn't work how you would expect. – 2017-02-04
  • 0
    @lulu Can I say that I fill instead of x (x/2) in your formula? And is the formula I was given correct? – 2017-02-04
  • 0
    Yes, you can just take $\frac x2$ in my formula, $\sum (-1)^n \frac {x^{2n}}n$. Or in your formula, for that matter. – 2017-02-04

1 Answers 1

1

If you want convergence on $[-2,2]$ then you should scale $x$ by a factor of two in your example. In order to get conditional convergence at $\{-2,2\}$ you may e.g. take: $$ f(x) = \ln(1+(x/2)^2) = \sum_{k\geq 1} \frac{(-1)^{k-1}}{4^k k} x^{2k} $$ (the point being that I have put the singularity into the complex plane).