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

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.