If $\forall n \in \mathbb{N}, a_n>0$ and $\lim_{n\rightarrow \infty }a_n=L>0$ does this imply that $\lim_{n\rightarrow \infty }\sqrt[n]{a_n}=1$?

Question

I/
I have a very simple question.
If $\forall n \in \mathbb{N}, a_n>0$ and $\lim_{n\rightarrow \infty }a_n=L>0$ does this imply that $\lim_{n\rightarrow \infty }\sqrt[n]{a_n}=\lim_{n\rightarrow \infty }\sqrt[n]{L}=1$?

II/ I ve answered yes to it and here is my proof.
Let define $b_{n\geq 2}=\frac{a_n}{a_{n-1}}\Rightarrow \lim_{n \to \infty }b_n=\lim_{n \to \infty }\frac{a_n}{a_{n-1}}=\frac{L}{L}=1$ (By limit arithmetic).
It is obvious too that $\forall n \geq 2 \in\mathbb{N}, b_n>0$ and that $\frac{a_n}{a_1}=b_2b_3...b_n \Rightarrow a_n=b_2b_3...b_na_1$
Now because we know that: $\lim_{n \to \infty }\sqrt[n]{b_2b_3...b_n}=1=\lim_{n \to \infty }b_n$ we can writte: $\lim_{n \to \infty }a_n^{1/n}=\lim_{n \to \infty }\sqrt[n]{b_2b_3...b_na_1}=\lim_{n \to \infty }\sqrt[n]{b_2b_3...b_n} \sqrt[n]{a_1}=\lim_{n \to \infty }\sqrt[n]{b_2b_3...b_n} \lim_{n \to \infty }\sqrt[n]{a_1}=1\cdot 1=1$. By arithmetic of limit and by knowing that:$\lim_{n \to \infty }\sqrt[n]{a_1}=1$ Hence $\lim_{n \to \infty }a_n^{1/n}=1=\lim_{n\rightarrow \infty }\sqrt[n]{L}$ Q.E.D.

I have two question:
1-Is my proof correct?
2-Does somebody know how to proove it by definition of limit (in the case this is true).

Thank you for your help.

Answer 1

Your proof is correct but uses a (relatively) sophisticated theorem to prove a basic result. Here is a simpler proof (alternatively, follow Martin R's advice).

If $a_n\to L>0$ then $(\ln a_n)$ is convergent hence bounded.

Therefore, $c_n:=\frac{\ln a_n}n\to0$ and $\sqrt[n]{a_n}=e^{c_n}\to e^0=1.$