I'm having some difficulty working through how to understand the ratio and root tests as they are presented in Rudin. I think the reason being is that limits are, in my experience, more straightforward to compute that $\limsup$s, and perhaps my understanding of $\limsup$/$\liminf$ isn't clear. Here are the definitions in Rudin, along with a sample problem. I would love clarification on what the definitions mean and practically how to use them.
Root Test:
1)Given $\sum a_n$ put $\alpha = \limsup_n\rightarrow \infty \sqrt[n]{|a_n|}$
then
a) if $\alpha <1$, the series converges.
b) if $\alpha >1$, the series converges
c) if $\alpha =1$, the test gives no information.
For this test, it seems identical to what I remember from calculus, replacing the limit with the limit supremum. What difference does this make, really? Is it saying that even if the limit doesn't strictly exist, as long as the terms are eventually strictly less than 1 that the series converges? Is it sufficient to simply demonstrate that $\exists N$ st $\forall~n>N, ~\sqrt[n]{|a_n|}<1$?
Ratio test
2) The series $\sum a_n$
a) converges if $\limsup n \rightarrow \infty \big{|}\frac{a_{n+1}}{a_n} \big{|}<1$
b) diverges if $\big{|}\frac{a_{n+1}}{a_n} \big{|} \geq 1$ for $n \geq n_0$ where $n_o$ is some fixed integer.
b) is confusing because it seems like this is essentially the definition of $\liminf$, but it doesn't say that. Am I understanding it correctly?
Practice problem
I would like to find the radius of convergence of $\sum n^3z^n$
It seems like the ratio test is easiest.
Set $\alpha = \limsup \big{|}\frac{(n+1)^3z^{n+1}}{n^3z^n} \big{|}$
= $\limsup \big{|}\frac{(n+1)^3z}{n^3} \big{|}$
=$\limsup \big{|}\frac{(n+1)^3}{n^3} \big{|} |z|$
Dropping absolute values because n is strictly positive
=$\limsup \big{(}\frac{n^3}{n^3}+\frac{3n^2}{n^3} + \frac{3n}{n^3} + \frac{1}{n^3} \big{)}|z|$
Now here I'm a little unsure how to proceed. By my reasoning, this means the $\limsup$ is $|z|$, as the limit of each of the terms involving $n$ is $1+0+0+0$, and it seems like if the quotients get arbitrarily close to 0, then the $\limsup= \liminf =0$. Is this correct? Is the radius of convergence the circle of radius 1 in that case?
Any tips or hints of clarifications would be appreciated. Thanks.
Best Answer
In order:
If $\limsup_{n \to \infty} \sqrt[n]{|a_n|} < 1$, then even if $\lim_{n \to \infty} |a_n|$ doesn't exist, we can find a series $(b_n)$ such that $\lim_{n \to \infty} |b_n| < 1$ and $(\forall n > n_0) |a_n| < |b_n|$. For example, write $L = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$, choose $\epsilon$ such that $L + \epsilon < 1$, and choose $n_0$ such that $(\forall n \geq n_0) \sqrt[n]{|a_n|} < L + \epsilon $. Then set $b_n = a_n$ for $n < n_0$ and $b_n = (L+\epsilon)^n$ for $n \geq n_0$. We can test the convergence of $(b_n)$ with the ordinary $\lim$ version of the root test, and if $(b_n)$ converges, then obviously $(a_n)$ must also converge. In this way, the $\limsup$ version of the test reduces to the $\lim$ version.
The test for convergence uses $\limsup$, whereas the test for divergence uses $\liminf$. You can find many series in which $\liminf_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$ and $\limsup_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1$, some of which diverge and some of which converge. Consider $(a_n), (b_n)$ defined recursively as $a_1 = b_1 = 1$ and then $$ \begin{align*} a_n &= \begin{cases}2 a_{n-1} & \text{$n$ is a power of 3} \\ \frac{1}{2} a_{n-1} & \text{otherwise} \end{cases} \\ b_n &= \begin{cases}\frac{1}{2} b_{n-1} & \text{$n$ is a power of 3} \\ 2 a_{n-1} & \text{otherwise} \end{cases} \end{align*}$$ Then although $\limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right| = \limsup_{n \to \infty} \left|\frac{b_{n+1}}{b_n} \right| = 2$ and $\liminf_{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right| = \liminf_{n \to \infty} \left|\frac{b_{n+1}}{b_n} \right| = \frac{1}{2}$, $\sum_{n=1}^\infty a_n$ converges and $\sum_{n=1}^\infty b_n$ diverges.