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

  • 1
    $\begingroup$ math.stackexchange.com/a/1708611/43949 $\endgroup$
    – angryavian
    Dec 3 '18 at 5:34
  • 2
    $\begingroup$ 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

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.


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.