Show that the power series $\sum_1^\infty a_nz^n$ and $\sum_1^\infty na_nz^n$ have the same radius of convergence

Show that the power series $$\sum_1^\infty a_nz^n$$ and $$\sum_1^\infty na_nz^n$$ have the same radius of convergence.

So I solved this problem using a method that I'm not sure is valid, and I just wanted to verify if this is rigorous enough and if not, try to understand the reason. Suppose $$\sum_1^\infty a_nz^n$$ has radius of convergence $$R$$, this implies $$\lim_{n\rightarrow \infty} |\frac{a_{n+1}}{a_n}|=\frac{1}{R}$$

Now using the ratio test for the second power series $$\lim_{n\rightarrow \infty} \left|\frac{(n+1)a_{n+1}}{na_n}\right|=\lim_{n\rightarrow \infty} \left|\left(1+\frac{1}{n}\right)\frac{a_{n+1}}{a_n}\right|=\frac{1}{R}$$

• This is not valid. You cannot conclude that $\lim |\frac {a_{n+1}} {a_n}|=\frac 1 R$. This limit need not even exist. May 3, 2020 at 13:20
• Oh I see, is there a simple example to when this occurs? May 3, 2020 at 13:25
• $a_n=2$ for $n$ even and $a_n=1$ for $n$ odd gives an example. May 3, 2020 at 13:27

Use the formula for radius of convergence and the fact that $$\lim n^{1/n}=1$$.
• @TigerAng The one derived from the root test, i.e. $R=\frac{1}{\lim \sup |c_n|^{1/n}}$. May 3, 2020 at 14:00
• $1/R=\lim \ sup |a_n|^{1/n}$. May 3, 2020 at 14:01
Compare reciprocals of radii of convergence: note that$$\lim_{n\to\infty}n^{1/n}=1\implies\limsup_{n\to\infty}|na_n|^{1/n}=\limsup_{n\to\infty}|a_n|^{1/n}.$$