If $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}=L$, then $\lim_{n\to\infty} (x_n)^{1/n}=L$ Let $x_n>0 \space \forall \space n \in N$
If $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}=L$, then $\lim_{n\to\infty} (x_n)^{1/n}=L$
This theorem is given in my book without proof. It looks very strange to me. Any suggestion about how to prove or derive it?
 A: Suppose $\lim_\limits{n\to\infty}\frac{x_{n+1}}{x_n}=L$. Then
$$\frac{1}{L}=\frac1{\lim_{n\to\infty}\frac{x_{n+1}}{x_n}}=\lim_{n\to\infty}\frac{x_n}{x_{n+1}}$$
by continuity of $\frac1x$ for $x>0$. We recognize $\lim_\limits{n\to\infty}\frac{x_n}{x_{n+1}}$ as the radius of convergence for the power series
$$\sum_{n=0}^\infty x_nx^n$$ which we will denote $R$. Similarly, we know that the radius of convergence for the above series is given as $\frac1{\limsup_\limits{n\to\infty}\sqrt[n]{x_n}}$ since $x_n>0$. Notice that
$$\frac1L=R=\frac1{\limsup_\limits{n\to\infty}\sqrt[n]{x_n}}=\frac1{\lim_\limits{n\to\infty}\sqrt[n]{x_n}}$$ in this case (due to our original assumption).
A: Here is a proof
from first principles.
If
$\lim_{n\to\infty} \frac{x_{n+1}}{x_n}=L
$,
then,
for any $c > 0$,
for all $n > n_0(c)$,
$L-c
\lt \frac{x_{n+1}}{x_n}
\lt L+c
$.
Therefore,
for any $N > n_0(c)$,
$(L-c)^{N-n_0(c)}
\lt \prod_{n=n_0(c)}^{N-1} \frac{x_{n+1}}{x_n}
=\dfrac{x_N}{x_{n_0(c)}}
\lt (L+c)^{N-n_0(c)}
$
so that
$x_{n_0(c)}(L-c)^{N-n_0(c)}
\lt x_N
\lt x_{n_0(c)}(L+c)^{N-n_0(c)}
$
or
$x_{n_0(c)}\dfrac{(L-c)^{N}}{(L-c)^{n_0(c)}}
\lt x_N
\lt x_{n_0(c)}\dfrac{(L+c)^{N}}{(L+c)^{n_0(c)}}
$.
Taking the $N$th root,
and rearranging a little,
$(L-c)\left(\dfrac{x_{n_0(c)}}{(L-c)^{n_0(c)}}\right)^{1/N}
\lt x_N^{1/N}
\lt (L+c)\left(\dfrac{x_{n_0(c)}}{(L+c)^{n_0(c)}}\right)^{1/N}
$.
Both
$\dfrac{x_{n_0(c)}}{(L-c)^{n_0(c)}}$
and
$\dfrac{x_{n_0(c)}}{(L+c)^{n_0(c)}}$
are independent of $N$,
so
$\lim_{N\to \infty} \left(\dfrac{x_{n_0(c)}}{(L\pm c)^{n_0(c)}}\right)^{1/N}
=1
$.
Therefore,
for large enough $N$,
$1-c 
\lt \left(\dfrac{x_{n_0(c)}}{(L\pm c)^{n_0(c)}}\right)^{1/N}
\lt 1+c
$,
so that
$(L-c)(1-c)
\lt x_N^{1/N}
\lt (L+c)(1+c)
$.
Now let $c \to 0$
and,
for each $c$,
let $N$ be sufficiently large
and we are done.
A: If $x_n>0 ,\forall n \in \mathbb{N}$ then $L>0$.

If $a_{n+1}-a_n \to a \in \mathbb{R}$ then $\frac{a_n}{n} \to a$

This is a corollary of Cesaro's theorem.
$\frac{x_{n+1}}{x_n} \to L$ thus $\log{\frac{x_{n+1}}{x_n}}=\log{x_{n+1}-\log{x_n}} \to \log{L}$
So from the corollary we have that $$\frac{\log{x_n}}{n} \to \log{L} \Rightarrow \log{\sqrt[n]{x_n}} \to \log{L} \Rightarrow \sqrt[n]{x_n} \to L$$
