# Radius of convergence and ratio test

My book says that given a power series $\sum_{n = 1}^\infty c_nz^n$ where the $c_n$ are complex the radius of convergence of the series is $\dfrac{1}{L}$ where $L = \lim \sup \sqrt[n]{|c_n|}$. So the radius of convergence is defined using the root test. Since we can also apply the ratio test, is it fair to say, that the radius of convergence is $\dfrac{1}{L}$ where $L = \lim \sup \bigg|\dfrac{c_{n+1}}{c_n}\bigg|$?

• You want $\lim \sup \frac{|c_{n+1}|}{|c_n|}$ for the ratio test. Mar 16 '13 at 3:38
• I think it's the same thing. $|a/b| = |a|/|b|$ for any complex $a, b$. Mar 16 '13 at 3:57
• The limit may be different, though. Mar 16 '13 at 3:59
• @LokiClock No, it can't. Mar 16 '13 at 4:04
• Okay. I think you're right. It's the same sequence, so it has the same limit. Mar 16 '13 at 4:37

Consider the series $$1+2z+3^2z^2+2^3z^3+3^4z^4+2^5z^5+3^6z^6+\cdots.$$ The required limsup of the $n$-th roots is $3$, and the radius of convergence is $\frac{1}{3}$.

Now look at the ratios $\dfrac{c_{n}}{c_{n+1}}$. You will notice they are very badly behaved, and tell us essentially nothing about the radius of convergence.

• Indeed, it tells us that $R\geq 0$... Nice example, +1. For some reason I could only come up with an example where $R>1/(\limsup |c_{n+1}|/|c_{n}|)=\liminf |c_n|/|c_{n+1}|>0$. Yours is the worst, hence the best possible. Mar 16 '13 at 14:19
• Now I wonder, in the case where the $c_n$'s are eventually nonzero. Does the formula $R=\liminf |c_n|/|c_{n+1}|$ hold if and only if the sequence $|c_n|/|c_{n+1}|$ converges? The if is easy, of course. But I can't see whether the only if holds or not. Any suggestion? Mar 16 '13 at 19:03
• @julien: For sure not immediately! Mar 16 '13 at 19:11

The ratio test is strictly weaker than the root test in the sense that if the ratio test gives an answer, then so does the root test and they are the same. However, as the other answers show, there are many series for which it gives no answer. Even though you are using $\limsup$ rather than $\lim$, the limit ratio may fail to be defined if the series contains infinitely many zeros, which make the corresponding ratios undefined, as in julien's answer.

In general, the ratio test is easier for common series because they involve algebraic operations for which computing the $n$th roots is difficult but for which the ratios can be computed pretty easily. But for theoretical purposes it is inferior.

The second formula does not hold. The ratio test only gives us bounds on the radius of convergence and only if applicable. To begin with, it requires the $$c_n$$'s be eventually nonzero. For instance $$\frac{1}{1-z^2}=\sum_{n\geq 0}z^{2n}.$$ has radius of convergence $$1$$. But $$c_{2n}=1$$ and $$c_{2n+1}=0$$. So the following is not defined: $$\not\exists\; \limsup \frac{|c_{n+1}|}{|c_{n}|}.$$ Now if the $$c_n$$'s are eventually nonzero, note that $$\limsup \frac{|c_{n+1}z^{n+1}|}{|c_nz^n|}=|z|\limsup \frac{|c_{n+1}|}{|c_{n}|}.$$ and $$\liminf \frac{|c_{n+1}z^{n+1}|}{|c_nz^n|}=|z|\liminf \frac{|c_{n+1}|}{|c_{n}|}.$$ A series $$\sum a_n$$ with $$a_n\geq 0$$ converges whenever $$\limsup a_{n+1}/a_n<1$$ and diverges whenever $$\liminf a_{n+1}/a_n>1$$. Also, note that $$\limsup a_n=\frac{1}{\liminf \frac{1}{a_n}}.$$ Hence our series converges if $$|z|\lt \frac{1}{\limsup\frac{|c_{n+1}|}{|c_n|}}=\liminf\frac{|c_n|}{|c_{n+1}|}$$ and diverges if $$|z|\gt \frac{1}{\liminf\frac{|c_{n+1}|}{|c_n|}}=\limsup\frac{|c_n|}{|c_{n+1}|}$$ It follows that the radius of convergence $$R$$ has $$\liminf \frac{|c_{n}|}{|c_{n+1}|}\leq R\leq \limsup \frac{|c_{n}|}{|c_{n+1}|}.$$ For an example where this is strict, take $$c_{2n}=2^n\cdot 3^n=6^n$$ and $$c_{2n+1}=2^n\cdot 3^{n+1}=3\cdot 6^n$$. Then $$\frac{1}{R}=\limsup \sqrt[n]{|c_n|}=\sqrt{6}<3=\limsup \frac{|c_{n+1}|}{|c_n|}.$$

Note that if $$\lim \frac{|c_{n+1}|}{|c_n|}$$ exists, then upper and lower limits coincide and we get a convenient formula $$R=\lim \frac{|c_{n}|}{|c_{n+1}|}.$$

• I'm not getting your first sentence. Which formula are you referring to, and what's incorrect? Mar 16 '13 at 4:02
• Are you sure? Wikipedia, Rudin, and pretty much everything says that the correct formula is the one I've given, so please provide a source. Mar 16 '13 at 4:08
• The ratio test as presented by Rudin says that the series $\sum a_n$ converges if $\lim \sup |a_{n+1}/a_n| < 1$. Mar 16 '13 at 4:11
• @user66960 This is correct. But here you have a power series...look at my edit. Mar 16 '13 at 4:12
• Let's apply your corrected version to the power series of $e^z$. $\lim \frac{1/n!}{1/(n+1)!} = \lim \frac{(n+1)!}{n!} = \infty$. Are you saying the radius of convergence of the power series of $e^z$ is $\frac{1}{\infty} = 0$? Please point where I'm going wrong. Mar 16 '13 at 4:31