limit of $\frac{a_{n+1}}{a_n}$

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

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

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.

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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.

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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