# 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 $x=x_1$

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.

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

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