I'm doing Problem II.9.3 in textbook Analysis I by Amann.
My attempt:
To prove that both series have the same radius of convergence. I need to show $$\limsup_{k \to \infty} \sqrt[k]{|a_k|} = \limsup_{k \to \infty} \sqrt[k]{ \left |(k+1)a_{k+1} \right|}$$
Because $$\limsup_{k \to \infty} \sqrt[k]{ \left |(k+1)\right|} = 1$$ I have to show $$\limsup_{k \to \infty} \sqrt[k]{|a_k|} = \limsup_{k \to \infty} \sqrt[k]{ \left |a_{k+1} \right|}$$
I'm stuck from here.
Could you please give me some hints to finish the proof? Thank you so much!
Update: Thanks to Botond's answer, I added the most important part here. All credits go to Botond.
Let $c = \limsup \sqrt[k]{|a_k|}$ and $d = \limsup \sqrt[k]{|a_{k+1}|}$.
For all $\varepsilon > 0$, there exists $N$ such that $\sqrt[k]{|a_k|} \le c + \varepsilon$ and $\sqrt[k]{|a_{k+1}|} \le d +\varepsilon$ for all $k \ge N$.
We have $|a_k| \le (c+\varepsilon)^{k}$, consequently $|a_{k+1}| \le (c+\varepsilon)^{k+1}$, and consequently $|a_{k+1}|^{1/k} \le (c+\varepsilon)^{(k+1)/k}$ for all $k \ge N$. Take the limit $k \to \infty$, we get $$d = \limsup \sqrt[k]{|a_{k+1}|} \le \limsup (c+\varepsilon)^{(k+1)/k} = c+\varepsilon ,\quad \varepsilon >0$$
Take the limit $\varepsilon \to 0$, we get $d \le c$.
We have $|a_{k+1}| \le (d+\varepsilon)^{k}$ and consequently $|a_{k+1}|^{1/(k+1)} \le (d+\varepsilon)^{k/(k+1)}$ for all $k \ge N$. Take the limit $k \to \infty$, we get $$c =\limsup \sqrt[k]{|a_{k}|} = \limsup \sqrt[k+1]{|a_{k+1}|} \le \limsup (d+\varepsilon)^{k/(k+1)} = d+\varepsilon ,\quad \varepsilon >0$$
Take the limit $\varepsilon \to 0$, we get $c \le d$.
As such, $c = d$. This completes the proof.
Best Answer
Continuing your work:
Let $c:=\lim\sup |a_n|^{1/n}$. Let $\varepsilon >0$ be given. Then we have that (for big enough $n$): $$|a_n|\leqslant (c+\varepsilon)^n$$ This means that $$|a_{n+1}|\leqslant (c+\varepsilon)^{n+1}$$ $$|a_{n+1}|^{1/n}\leqslant (c+\varepsilon)^{1+1/n}$$ I think you can finish this direction, and work out the other direction as well.