Show that this limit is equal to $\liminf a_{n}^{1/n}$ for positive terms.

Show that if $a_{n}$ is a sequence of positive terms such that $\lim\limits_{n\to\infty} (a_{n+1}/a_n)$ exists, then this limit is equal to $\liminf\limits_{n\to\infty} a_n^{1/n}$.

I am not event sure where to start from, any help would be much appreciated.

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Consider $a_n=\frac{a_n}{a_n-1}\frac{a_n-1}{a_n-2}....\frac{a_2}{a_1}$, and I think you can complete the following. – 89085731 Mar 4 '12 at 2:14
This follows also from the inequality $\liminf \frac{a_{n+1}}{a_n}\leq \liminf \sqrt[n]{a_n} \le \limsup \sqrt[n]{a_n}\leq \limsup \frac{a_{n+1}}{a_n}$, which is true for positive sequences. See this question, this answer and this answer. – Martin Sleziak Jun 22 '12 at 8:26

Proof Let us define new sequence $(b_n)_{n=1}^{\infty}$ with $b_1=a_1$ and $b_n=\frac{a_{n+1}}{a_n}$ for all $n>1$.

Now, we consider the following as known,

If $(x_n)_{n=1}^{\infty}$ is a sequence of positive numbers and $\lim x_n=L$, then,

$$\lim\sqrt[n]{x_1x_2...x_n}=L$$.

With that statement, we will continue and say,

$$\lim\sqrt[n]{b_1b_2...b_n}=\lim b_n=\lim \frac{a_{n+1}}{a_n}$$

And its clear that,

$$b_1b_2...b_n=a_{n+1}$$

Hence,

$$\lim \sqrt[n]{a_{n+1}}=\lim\frac{a_{n+1}}{a_n}$$

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I make the same proof with you,but I don't know how to apply inf. – 89085731 Mar 4 '12 at 2:30
@ Gingerjin: Theorem: lim a_n=l iff liminf a_n=limsup a_n=l – Salech Alhasov Mar 4 '12 at 2:34
@SalechAlhasov I could n't follow why if we assume that $\lim x_n=L$, then, $$\lim\sqrt[n]{x_1x_2...x_n}=L$$ in all cases. Although it sounds very reasonable but i would be great full if you could shed some more light on it. – Comic Book Guy Mar 4 '12 at 2:39
Yes. First read here, link, then prove that if $\lim a_n=L$ then $\lim\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+...\frac{1}{a_n}}=L$, afterwards use AM-GM-HM and sandwich theorem to conclude the wanted. – Salech Alhasov Mar 4 '12 at 2:51

The typical textbook proof goes as follows.

If the $\lim \frac{a_{n+1}}{a_n} = q \gt 0$, then given an $\epsilon \gt 0$ such that $q \gt \epsilon$, there is some $n_0$ such that for all $n \ge n_0$

$$q - \epsilon \lt \frac{a_{n+1}}{a_n} \lt q+\epsilon$$

Multiplying gives us

$$a_{n_0}(q - \epsilon)^{n-n_0} \lt a_{n} \lt (q+\epsilon)^{n-n_0} a_{n_0}$$

and so

$$a_{n_0}^{1/n}(q - \epsilon)^{1-n_0/n} \lt a_{n}^{1/n} \lt (q+\epsilon)^{1-n_0/n} a_{n_0}^{1/n_0}$$

And thus (by taking the limit as $n \to \infty$)

$$q - \epsilon \le \liminf (a_n)^{1/n} \le \limsup (a_n)^{1/n} \le q + \epsilon$$

Since $\epsilon$ was arbitrary, we have that $q = \lim a_n^{1/n}$.

The case $q=0$, we replace left hand side by $0$, and the proof carries through.

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(+1) Such a useful theorem with such an easy proof, it should be more well known. For some reason, people seem to think it is a hard theorem. – Ragib Zaman Mar 4 '12 at 2:53
@RagibZaman: So true! +1 to your answer there. – Aryabhata Mar 4 '12 at 2:55
@Aryabhata. I'm a little confused. What did you multiply by after you had: $q−\epsilon < \frac{a_{n+1}}{an} < q+ \epsilon$ ? – user66360 Dec 1 '13 at 18:24
@kbball: Take the inequality for $n = n_0$, $n = n_0 + 1$, and so on till $n$ and then multiply all. – Aryabhata Dec 1 '13 at 18:48

I saw this proof today and thought it's a nice one:

Let $a_n\ge 0$, $\lim\limits_{n \to \infty}a_n=L$. So there are 2 options:

(1) $L>0$: $$\lim\limits_{n \to \infty}a_n=L \iff \lim\limits_{n \to \infty}\frac{1}{a_n}=\frac{1}{L}$$ Using Cauchy's Inequality Of Arithmetic And Geometric Means we get:

$$\frac{n}{a_1^{-1}+\dots+a_n^{-1}}\le\sqrt[n]{a_1\cdots a_n}\le \frac{a_1+ \cdots + a_n}{n}$$ Applying Cesaro Theorem on $a_n$, notice that RHS $\mathop{\to_{n \to \infty}} L$ , and that by applying Cesaro Thm on $1/a_n$, LHS$\mathop{\to_{n \to \infty}} \frac{1}{1/L}=L$ . And so from the squeeze thm: $$\lim\limits_{n \to \infty}\sqrt[n]{a_1\cdots a_n}=L$$

(2) $L=0$:
$$0\le\sqrt[n]{a_1\cdots a_n}\le \frac{a_1+ \cdots + a_n}{n}$$

$$\Longrightarrow\lim\limits_{n \to \infty}\sqrt[n]{a_1\cdots a_n}=0=L$$

Now, define:

$$b_n = \begin{cases}{a_1} &{n=1}\\\\ {\frac{a_n}{a_{n-1}}} &{n>1}\end{cases}$$ and assume $\lim\limits_{n \to \infty}b_n=B$.

Applying the above result on $b_n$ we get: $$\frac{n}{b_1^{-1}+\dots+b_n^{-1}}\le\sqrt[n]{b_1\cdots b_n}\le \frac{b_1+ \cdots + b_n}{n}$$ $$\frac{n}{b_1^{-1}+\dots+b_n^{-1}}\le\sqrt[n]{a_1\cdot (a_2/a_1)\cdots (a_n/a_{n-1})}\le \frac{b_1+ \cdots + b_n}{n}$$ $$\frac{n}{b_1^{-1}+\dots+b_n^{-1}}\le\sqrt[n]{a_n}\le \frac{b_1+ \cdots + b_n}{n}$$ $$\Longrightarrow\lim\limits_{n \to \infty}\sqrt[n]{a_n}=B$$

So we can conclude that if $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}$ exists and equal to $L$, then $\lim\limits_{n \to \infty}\sqrt[n]{a_n}=L$.

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Hint:

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

$$(a_{n+1}/a_n)= e^{\ln (a_{n+1}) -\ln(a_n)} \,.$$

Try to prove instead the equality of the corresponding exponents limits..

P.S. Are you familiar with Stolz Cezaro or Cezaro means?

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Is inf here necessary？ – 89085731 Mar 4 '12 at 2:21
No, but i will look them up on wikipedia and try to apply. Would post back if i run into trouble. – Comic Book Guy Mar 4 '12 at 2:22
@Gingerjin You are right, is not needed. The fact that it is there is probably relevant to where this comes from, might be a part of a proof of a theorem. – N. S. Mar 4 '12 at 2:33
One more comment: it looks to me that if one can prove this result with linminf, one gets for free that the limit exists: apply this result for $b_n = \frac{1}{a_n}$ and get that limsup $a_n$= liminf $a_n$. – N. S. Mar 4 '12 at 2:35
This argument might fail when $a_n \to 0$. – Aryabhata Mar 4 '12 at 2:38