# Thread: [SOLVED] A demonstration regarding the radius of convergence of an infinite series

1. ## [SOLVED] A demonstration regarding the radius of convergence of an infinite series

Given the power series $\sum _{n=0}^{\infty} a_n (z-z_0)^n$, prove that if $\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ 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 $L= \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ exists and if $L<1$ then the series converges. If $L=1$ then we can't conclude by the test. If $L>1$ then the series diverges.

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

Thus I must show that $\frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$.
I first assume that $L<1$, in other words that $|a_n|<|a_{n+1}|$.
I've tried to continue the proof ( by showing $\frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ using the fact that $|a_n|<|a_{n+1}|$ ) but without success.

Can you help me a little bit? Thanks.

2. Originally Posted by arbolis
Given the power series $\sum _{n=0}^{\infty} a_n (z-z_0)^n$, prove that if $\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ 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 $\color{red}L= \lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ exists and if $\color{red}L<1$ then the series converges. If $\color{red}L=1$ then we can't conclude by the test. If $\color{red}L>1$ then the series diverges.

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

Thus I must show that $\frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$.
I first assume that $L<1$, in other words that $|a_n|<|a_{n+1}|$.
I've tried to continue the proof ( by showing $\frac{1}{\lim _{n \to \infty} |a_n|^{\frac{1}{n}} }=\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$ using the fact that $|a_n|<|a_{n+1}|$ ) but without success.

Can you help me a little bit? Thanks.
The sentence in red is a correct statement about the series $\sum _{n=0}^{\infty} a_n$. But you are trying to apply it to the series $\sum _{n=0}^{\infty} a_n (z-z_0)^n$, in which the n'th term is not $a_n$ but $a_n(z-z_0)^n$. So what the ratio test is telling you is that if $L= \lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right|$ exists and if $L<1$ then the series converges. If $L=1$ then we can't conclude by the test. If $L>1$ then the series diverges.

But $\lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right| = \frac1{|z-z_0|}\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$. From that, you can conclude directly that the power series converges if $|z-z_0|<\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$, and diverges if $|z-z_0|>\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$.

3. Originally Posted by Opalg
The sentence in red is a correct statement about the series $\sum _{n=0}^{\infty} a_n$. But you are trying to apply it to the series $\sum _{n=0}^{\infty} a_n (z-z_0)^n$, in which the n'th term is not $a_n$ but $a_n(z-z_0)^n$. So what the ratio test is telling you is that if $L= \lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right|$ exists and if $L<1$ then the series converges. If $L=1$ then we can't conclude by the test. If $L>1$ then the series diverges.

But $\lim _{n \to \infty} \left| \frac{a_n(z-z_0)^n}{a_{n+1}(z-z_0)^{n+1}} \right| = \frac1{|z-z_0|}\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$. From that, you can conclude directly that the power series converges if $|z-z_0|<\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$, and diverges if $|z-z_0|>\lim _{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$.
Oh thank you very much. That was very useful!