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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.2012-03-04
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    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](http://math.stackexchange.com/questions/69386/inequality-involving-limsup-and-liminf), [this answer](http://math.stackexchange.com/a/76800/) and [this answer](http://math.stackexchange.com/questions/28476/finding-the-limit-of-frac-n-sqrtnn/28487#28487).2012-06-22

4 Answers 4

1

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?2012-03-04
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    No, but i will look them up on wikipedia and try to apply. Would post back if i run into trouble.2012-03-04
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    @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.2012-03-04
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    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 $a_n$.2012-03-04
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    This argument might fail when $a_n \to 0$.2012-03-04
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    @Aryabhata Stolz Cezaro only requires that the denominator is increasing and bounded, there is no restriction on the numerator.2012-03-04
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    I was talking about the statement: "try to prove equality of the exponent limits".2012-03-04
6

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](http://math.stackexchange.com/questions/115143/prove-that-varlimsup-n-rightarrow-inftyu-n-frac1n-1-where-u-n/115245#115245) theorem.2012-03-04
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    @RagibZaman: So true! +1 to your answer there.2012-03-04
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    @Aryabhata. I'm a little confused. What did you multiply by after you had: $q−\epsilon < \frac{a_{n+1}}{an} < q+ \epsilon$ ?2013-12-01
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    @kbball: Take the inequality for $n = n_0$, $n = n_0 + 1$, and so on till $n$ and then multiply all.2013-12-01
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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.2012-03-04
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    @ Gingerjin: Theorem: lim a_n=l iff liminf a_n=limsup a_n=l2012-03-04
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    @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.2012-03-04
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    Yes. First read here, [link](http://www.ee.columbia.edu/~vittorio/CesaroMeans.pdf), 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.2012-03-04
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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 $\lim\limits_{n \to \infty}\sqrt[n]{a_n}=L$.