Is $\limsup_{n\rightarrow\infty}\frac{c_{n+1}}{c_{n}}=\limsup_{n\rightarrow\infty}\sqrt[n]{c_{n}}$ for a positive sequence $(c_{n})_{n\geq 1}$?

Question

I was going through Theorem 3.39 from Baby Rudin page no 69. I will restate the theorem,

Given the power series $\sum_{n\geq0}c_{n}z^{n}$, put $\alpha=\limsup_{n\rightarrow\infty}\sqrt[n]{|c_{n}|}$ and $R=\frac{1}{\alpha}$(If $\alpha=0$, $R=\infty$ and vice-versa). Then $\sum_{n\geq0}c_{n}z^{n}$ converges if $|z|<R$ and diverges if $|z|>R$.

In the proof of the theorem, the root test is applied to individual terms in the power series. I tried the same using ratio test and the expression of reciprocal of the radius of convergence becomes $\alpha = \limsup_{n\to\infty}\frac{|c_{n+1}|}{|c_{n}|}$. Then using the inequality $\limsup_{n\to\infty}\frac{|c_{n+1}|}{|c_{n}|}\geq\limsup_{n\rightarrow\infty}\sqrt[n]{|c_{n}|}$, if it is followed strictly, there will be two distinct radius of convergence(let $R_{1}$ and $R_{2}$) and this cannot be the case, since the series will converge and diverge in an annular region with radius $R_{1}$ and $R_{2}$, by ratio test and root test.

So, the only possible scenario can be the two are equal, implying equality holds in that inequality, i.e. $\limsup_{n\to\infty}\frac{|c_{n+1}|}{|c_{n}|}=\limsup_{n\rightarrow\infty}\sqrt[n]{|c_{n}|}$. But this looks so absurd since it makes the inequality totally meaningless. I am unable to figure out where is the gap in my thinking. Any help will be appreciated.

Answer 1

If $(c_n)_{n\in\Bbb N}$ is the sequence$$1,1,2,2,4,4,8,8,\ldots$$then $\lim_{n\to\infty}\sqrt[n]{c_n}=\sqrt2$ and, in particular, $\limsup_n\sqrt[n]{c_n}=\sqrt2$, but $\limsup_n\frac{c_{n+1}}{c_n}=2$ (and $\liminf_n\frac{c_{n+1}}{c_n}=1$).

Answer 2

1. The ratio test with the resulting limit being $r$ correctly yields $R=\frac{1}{r}$ if the relevant limit exists, but there is no guarantee that it does. The reason for this is basically that a comparison to the geometric series with common ratio $r$ results in tight bounds except when $|z|=R$.

2. The ratio test with the resulting limsup (resp. liminf) being $r$ in general gives only $R \geq \frac{1}{r}$ (resp. $R \leq \frac{1}{r}$). The reason for this is basically that a comparison to the geometric series can be extremely far from tight, due to infinitely many consecutive ratios being significantly smaller than the limsup (resp. larger than the liminf).

3. The root test with a limsup gives the optimal radius of convergence.

A simple example for #2 with all nonzero coefficients is a sequence of coefficients where the odd coefficients are "way bigger" than the even coefficients, like $a_{2n}=2^{-n}$ but $a_{2n+1}=1$. Then $\frac{a_{2n}}{a_{2n-1}}=2^{-n}$ while $\frac{a_{2n+1}}{a_{2n}}=2^n$, so the limsup is infinity and the liminf is zero. So the ratio test with a limsup just tells you $R \geq 0$ and the ratio test with a liminf tells you $R \leq \infty$, both useless. Meanwhile the radius of convergence is actually $1$.