Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Today my lecturer put up on the board that:

If $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}$ exists and $a_n>0$ then

$\displaystyle \limsup\limits_{n\to\infty}\left(a_n^{\frac{1}{n}}\right)=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$

however I am not sure why this is true, can somebody give me a hint or something as to how to go about proving this.

thanks for any help

share|cite|improve this question
$a_n^{\frac{1}{n}}= e^{\frac{ \ln (a_n)}{n}} \,. $ Do you know how to calculate the limit in the exponent? ;) – N. S. Oct 28 '11 at 18:55
Why the limsup? – Did Oct 28 '11 at 18:59
Also, this would be a nice thing to ask the lecturer. – GEdgar Oct 28 '11 at 19:00
You left out the assumption $a_n > 0$. Hint: if $a_{n+1}/a_n \ge K$ for all $n \ge N$ then $a_n/a_N \ge \ldots$. Similarly for $\le$. – Robert Israel Oct 28 '11 at 19:04
Further hint: if $a_n/a_N \le \ldots$, then $a_n^{1/n} \le \ldots$. Note that for any $c > 0$, $c^{1/n} \to c^0 = 1$ as $n \to \infty$. – Robert Israel Oct 28 '11 at 21:17
up vote 6 down vote accepted

In fact, the stronger statement is as follows:

Theorem: Let $\{c_n\}$ be any sequence in $\mathbb{R}^+$. Then, $\displaystyle \underline{\lim}\frac{c_{n+1}}{c_n}\leq \underline{\lim}\sqrt[n]{c_n}$ and $\displaystyle \overline{\lim}\sqrt[n]{c_n}\leq \overline{\lim}\frac{c_{n+1}}{c_n}$.

So, with this, if we assume that $\displaystyle \lim\frac{c_{n+1}}{c_n}$ exists then we have that $\displaystyle \overline{\lim}\sqrt[n]{c_n}\leq\overline{\lim}\frac{c_{n+1}}{c_n}=\underline{\lim}\frac{c_{n+1}}{c_n}\leq \underline{\lim}\sqrt[n]{c_n}$ from where it easily follows that $\overline{\lim}\sqrt[n]{c_n}=\underline{\lim}\sqrt[n]{c_n}$ and so $\lim \sqrt[n]{c_n}$ exists and, in fact, it's also clear it must be equal to $\displaystyle \lim\frac{c_{n+1}}{c_n}$. A proof of this fact can be found on page 68 of Rudin's Principles of Mathematical Analysis. I assume you have access to this (very well-known) book--if not say so and I shall give an outline of the proof.

share|cite|improve this answer
I don't at the moment and will not for about a week, would it be possible for you to outline the proof. (I will have my book back in a while so if you are busy its not a problem) Thanks very much for the response. – hmmmm Oct 28 '11 at 22:33
Wait no need, I have it. Thanks very much – hmmmm Oct 28 '11 at 23:15
The same result was used in this answer. It can be derived from this form of Stolz-Cesaro theorem. – Martin Sleziak Oct 30 '11 at 11:39

As I mentioned in my comment,

$$a_n^{\frac{1}{n}}= e^{\frac{ \ln (a_n)}{n}} $$

Now, if the limit

$$\lim_{n \to \infty} \frac{\ln (a_{n+1})-\ln (a_n)}{(n+1)-n}= \lim_{n \to \infty} \ln \left( \frac{a_{n+1}}{a_n} \right) $$ exists then by Stolz Cezaro the limit $$\lim_{n \to \infty} \frac{ \ln (a_n)}{n}$$ exists and

$$\lim_{n \to \infty}\frac{ \ln (a_n)}{n}= \lim_{n \to \infty} \ln \left(\frac{a_n+1}{a_n}\right) $$

The Theorem mentioned in the other post also follows from the stronger version of Stolz Cezaro by exactly the same reasoning.

share|cite|improve this answer
Im a bit confused here have you not shown that $\displaystyle \mbox{lim} (a_n)^{(\frac{1}{n})}=\mbox{lim} \frac{a_{n+1}}{a_n}$? – hmmmm Oct 30 '11 at 23:43
@hmmmm Yep, there is no need for sup limit here. If $\lim \frac{a_{n+1}}{a_n}$ exists then $\lim_n \sqrt[n]{a_n}$ automatically exists and is the same, and this is what I proven... The converse is not true though..... – N. S. Jun 2 '12 at 10:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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