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**arbolis** Given the power series $\displaystyle \sum _{n=0}^{\infty} a_n (z-z_0)^n$, prove that if $\displaystyle \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ exists and is finite or infinite, then it is equal to the radius of convergence of the series.

Suggestion : Use the ratio test.

My attempt : The ratio test states that if $\displaystyle \color{red}L= \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ exists and if $\displaystyle \color{red}L<1$ then the series converges. If $\displaystyle \color{red}L=1$ then we can't conclude by the test. If $\displaystyle \color{red}L>1$ then the series diverges.

The radius of convergence of the power series is $\displaystyle R=\frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }$.

Thus I must show that $\displaystyle \frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$.

I first assume that $\displaystyle L<1$, in other words that $\displaystyle |a_n|<|a_{n+1}|$.

I've tried to continue the proof ( by showing $\displaystyle \frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ using the fact that $\displaystyle |a_n|<|a_{n+1}|$ ) but without success.

Can you help me a little bit? Thanks.