Prove or disprove:

if $\sum_{n=0}^\infty a_n x^n$ converges at $x=x_1$ , then $\sum_{n=0}^\infty n\cdot a_n \cdot x^{n-1}$ converges at $x=x_1$

I am quite new to this material (and taylor series especially).

I am pretty sure, that if I differentiate a power series, the radius of convergence stays the same, but:

  1. I'm not sure why.

  2. if $R=x_1$ (The radius of convergence), and it converges in the original series, I don't think it still holds for the differentiate.

Would love some guidelines.

  • $\begingroup$ Hint: suppose $x_1=1$. $\endgroup$ – lulu Jun 17 '17 at 10:33
  • 1
    $\begingroup$ The $\limsup \sqrt[n]{|n a_n|} = \limsup \sqrt[n]{|a_n|}$ should help (because $\sqrt[n]{n} \to 1$). $\endgroup$ – Sil Jun 17 '17 at 10:39

You are right about the radius of convergence. That is why the case to explore is the boundary of the disc.

Try $$\sum_n\frac{x^n}{n^2}$$ at $x=1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.