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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.

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  • $\begingroup$ Hint: suppose $x_1=1$. $\endgroup$ – lulu Jun 17 '17 at 10:33
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    $\begingroup$ 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
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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 $x=1$.

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