I am trying to prove that $\displaystyle \lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = L$ implies $\displaystyle \lim_{k \to \infty} {a_{k}}^{1/k} = L$
This what I did:
By the limit definition, if $\displaystyle \lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = L $ then $\left | \dfrac{a_{n+1}}{a_n}- L \right | < \epsilon$ for all $n \geq m$. Therefore, it is also true that
$L - \epsilon < \dfrac{a_{m+1}}{a_m} < L + \epsilon$
$L - \epsilon < \dfrac{a_{m+2}}{a_{m+1}} < L + \epsilon$
$L - \epsilon < \dfrac{a_{m+4}}{a_{m+3}} < L + \epsilon$
$\quad \vdots$
$L - \epsilon < \dfrac{a_{k}}{a_{k-1}} < L + \epsilon$
Multiplying all inequalities and simplyfing:
$(L - \epsilon)^{k-m} < \dfrac{a_{k}}{a_{m}} < (L + \epsilon)^{k-m}$
and applying the k-root:
$(L - \epsilon)^{1-m/k} < \dfrac{{a_{k}}^{1/k}}{{a_{m}}^{1/k}} < (L + \epsilon)^{1-m/k}$
Since $m$ is fixed, taking the limit $k \to \infty$ we get
$ L - \epsilon < \displaystyle \lim_{k \to \infty}{a_{k}}^{1/k} < (L + \epsilon) $
My problem is I don't know how to justify from the last step that $\displaystyle \lim_{k \to \infty}{a_{k}}^{1/k} = L$. I asked myself that if I have $L-\epsilon < \lim a(n) < L+\epsilon$ for all $\epsilon > 0$ would imply $\lim a(n) = L$, but I couldn't get a good answer.
Best regards!
In fact, you are assuming the limit exists without proof.
However, we have for all $\epsilon > 0$,
$$L- \epsilon \leqslant \liminf_{ \,k \to \infty} \,a_k^{1/k} \leqslant \limsup_{\,k \to \infty}\,a_k^{1/k} \leqslant L+\epsilon$$.
This proves (since assuming otherwise leads to a contradiction)
$$L = \liminf_{k \to \infty} \,a_k^{1/k} = \limsup_{k \to \infty}\,a_k^{1/k} = \lim_{k \to \infty} a_k^{1/k}$$