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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

  • 3
    $a_n^{\frac{1}{n}}= e^{\frac{ \ln (a_n)}{n}} \,. $ Do you know how to calculate the limit in the exponent? ;)2011-10-28
  • 0
    Why the limsup?2011-10-28
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    Also, this would be a nice thing to ask the lecturer.2011-10-28
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    @Didier: I *think* it was the limsup; the OP had `\lim` followed by `\mathrm{sup}`; I guess it could have been the limit of the suprema instead...2011-10-28
  • 3
    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 $\le$.2011-10-28
  • 0
    It has to be limsup. There is a simple alternating example to show that lim does not always exists. (Only the other way round.)2011-10-28
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    Yeah it was meant to be limsup, @ArturoMagidin sorry about that I must have made a typo thanks for the edit2011-10-28
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    @abatkai, *There is a simple alternating example* showing that $a_{n+1}/a_n$ converges but $(a_n)^{1/n}$ diverges? Is there? Most curious to see your example, I must say.2011-10-28
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    ok sorry but imnot really getting anywhere with this sorry, could I get some more help or if it should be obvious from what you have said (and I just can't get it) could I have a solution to look at, thanks for all the help2011-10-28
  • 0
    @user, right, the hypothesis that $a_n$ is positive is missing.2011-10-28
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    Realised that, that's why I deleted my comment.2011-10-28
  • 1
    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 $n \to \infty$.2011-10-28

2 Answers 2

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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.

  • 0
    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.2011-10-28
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    Wait no need, I have it. Thanks very much2011-10-28
  • 1
    The same result was used in [this answer](http://math.stackexchange.com/questions/28476/finding-the-limit-of-n-sqrtnn/28487#28487). It can be derived from [this form](http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup/73709#73709) of Stolz-Cesaro theorem.2011-10-30
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    I presume $\overline\lim$ means $\limsup$?2018-10-03
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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.

  • 0
    Im a bit confused here have you not shown that $\displaystyle \mbox{lim} (a_n)^{(\frac{1}{n})}=\mbox{lim} \frac{a_{n+1}}{a_n}$?2011-10-30
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    @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.....2012-06-02