# Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$

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

• $a_n^{\frac{1}{n}}= e^{\frac{ \ln (a_n)}{n}} \,.$ Do you know how to calculate the limit in the exponent? ;) 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. 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 $\le$. 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 $n \to \infty$. 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.

• 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. Oct 28 '11 at 22:33
• Wait no need, I have it. Thanks very much 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. Oct 30 '11 at 11:39
• I presume $\overline\lim$ means $\limsup$? Oct 3 '18 at 5:15
• Is there an example where the middle equality holds but the other two (or at least one) is strict?
– JKEG
Nov 27 '19 at 2:27

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.

• Im a bit confused here have you not shown that $\displaystyle \mbox{lim} (a_n)^{(\frac{1}{n})}=\mbox{lim} \frac{a_{n+1}}{a_n}$? 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..... Jun 2 '12 at 10:03

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.