# Theorem 3.37 in Baby Rudin: Any Counter-Examples In The Other Case?

Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:

For any sequence $$\left\{ c_n \right\}$$ of positive numbers, $$\lim\inf_{n\to\infty} \frac{c_{n+1}}{c_n} \leq \lim\inf_{n\to\infty} \sqrt[n]{c_n} \leq \lim\sup_{n\to\infty} \sqrt[n]{c_n} \leq \lim\sup_{n\to\infty} \frac{c_{n+1}}{c_n}.$$

I think I fully understand the proof by Rudin.

From Theorem 3.37 in Baby Rudin, we can also conclude that following:

For any sequence $$\left\{ c_n \right\}$$ of positive numbers, if the sequence $$\left\{ \frac{c_{n+1}}{c_n} \right\}$$ converges in $$\mathbb{R}$$, then so does the sequence $$\left\{ \sqrt[n]{c_n} \right\}$$, and then the two limits are equal.

Am I right?

However, I'm unable to figure out the proof of or come up with any counter-examples to the following:

Suppose that $$\left\{ c_n \right\}$$ is a sequence of positive real numbers such that the sequence $$\left\{ \sqrt[n]{c_n} \right\}$$ converges in $$\mathbb{R}$$. Then so does the sequence $$\left\{ \frac{c_{n+1}}{c_n} \right\}$$.

What if the sequence $$\left\{ \sqrt[n]{c_n} \right\}$$ converges in $$\mathbb{R} \cup \{ \pm \infty \}$$? Does the sequence $$\left\{ \frac{c_{n+1}}{c_n} \right\}$$ then also converge in $$\mathbb{R} \cup \{ \pm \infty \}$$?

• math.stackexchange.com/a/1708611/43949 Dec 3 '18 at 5:34
• Think as $\sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $\log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges… Dec 3 '18 at 5:42

On the other hand, consider the sequence$$1,1,\frac12,\frac12,\frac14,\frac14,\ldots$$In this case, $$\lim_{n\to\infty}\sqrt[n]{c_n}=\dfrac1{\sqrt2}$$, but $$\lim_{n\to\infty}\dfrac{c_{n+1}}{c_n}$$ doesn't exist.
Hint $$1,2,1,2,1,2,1,2,...$$
The sequence $$\sqrt[n]{c_n}$$ converges since $$\lim_n \sqrt[n]{1}=\lim_n \sqrt[n]{2}=1$$