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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 $\mathbb{R} \cup \{ \pm \infty \}$?

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    $\begingroup$ math.stackexchange.com/a/1708611/43949 $\endgroup$
    – angryavian
    Dec 3 '18 at 5:34
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    $\begingroup$ 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… $\endgroup$
    – Idéophage
    Dec 3 '18 at 5:42
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Concerning the first question: yes, you are right.

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.

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Hint $$1,2,1,2,1,2,1,2,...$$

The sequence $\sqrt[n]{c_n}$ converges since $$\lim_n \sqrt[n]{1}=\lim_n \sqrt[n]{2}=1$$

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