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I have the power seires $$\sum_{k=0}^\infty{a_kz^k}$$ with $z, a_k \in \mathbb{C}$.
Say this series has radius of convergence $R$. I want to show that then $$\sum_{k=1}^\infty{a_k k z^{k-1}}$$ has the same radius of convergence $R$.

Can someone give me a hint how to approach this? I tried to apply ratio/root test but it does not seem to help much.

Thanks!

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  • $\begingroup$ Show $\limsup \sqrt[k]{a_k} = \limsup \sqrt[k-1]{ka_k}$. $\endgroup$ – Thomas Andrews Dec 3 '15 at 13:40
  • $\begingroup$ I actually thought about that but then I got stuck with $\text{lim sup} \sqrt[k-1]{k} ~~\sqrt[k-1]{a_k}$. I don't think its valid just to say: Since $\sqrt[k]{k} \to 1$ the rest follows trivally. Is it possible to split the first term off (since I don't know anything about $a_k$)? $\endgroup$ – elfeck Dec 3 '15 at 13:59
  • $\begingroup$ If $b_k\to 1$ then $\limsup b_kc_k = \limsup c_k$. So you really only need to show that $\limsup \sqrt[k-1]{a_k}=\limsup\sqrt[k]{a_k}$. $\endgroup$ – Thomas Andrews Dec 3 '15 at 14:04
  • $\begingroup$ I've been trying for a bit now but I can't seem to find anything helpful to show this. $\endgroup$ – elfeck Dec 3 '15 at 14:29
  • $\begingroup$ $$|a_k|^{1/(k-1)}=\bigl(|a_k|^{1/k}\bigr)^{k/(k-1)}$$ $\endgroup$ – Julián Aguirre Dec 3 '15 at 14:35

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