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.
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.
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,
$$ \lim \sqrt[n]{a_{n+1}}=\lim\frac{a_{n+1}}{a_n}$$
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 $B$, then $\lim\limits_{n \to \infty}\sqrt[n]{a_n}=B$.
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?
