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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

Watson
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hmmmm
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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
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    Also, this would be a nice thing to ask the lecturer. – GEdgar Oct 28 '11 at 19:00
  • @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... – Arturo Magidin Oct 28 '11 at 19:03
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    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
  • It has to be limsup. There is a simple alternating example to show that lim does not always exists. (Only the other way round.) – abatkai Oct 28 '11 at 19:07
  • Yeah it was meant to be limsup, @ArturoMagidin sorry about that I must have made a typo thanks for the edit – hmmmm Oct 28 '11 at 19:08
  • @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. – Did Oct 28 '11 at 19:31
  • 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 help – hmmmm Oct 28 '11 at 19:44
  • @user, right, the hypothesis that $a_n$ is positive is missing. – Did Oct 28 '11 at 19:50
  • Realised that, that's why I deleted my comment. – N. S. Oct 28 '11 at 20:04
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    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

3 Answers3

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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.

Pedro
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Alex Youcis
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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.

Mark Viola
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N. S.
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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
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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..... – N. S. Jun 02 '12 at 10:03
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The equality of $\limsup_{n\to\infty}a_n^{1/n}$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ is actually something you already learned in Calculus 1, just in disguise. If we consider $\sum_{n=1}^{\infty}a_n x^n$, then the radius of convergence is equal to the first limit by the root test and it is also equal to the second limit by the ratio test, and thus both limits are equal.

Milo Moses
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