How to show that $\lim_{n \to \infty} a_n^{1/n} = l$?

Question

Suppose that $a_n > 0$, $n \in \mathbb{N}$. Suppose that $$\lim_{n \to \infty} a_{n+1}/a_n =l$$.

How to show that

$$\lim_{n \to \infty} a_n^{1/n} = l \;?$$

My solution: let $$b_n = a_{n+1}/a_n$$

Then

$$b_1 b_2 \cdots b_{n-1} = a_{n}/a_1$$.

Do we have $\lim_{n \to \infty} (b_1 \cdots b_{n-1})^{1/n} = l\;?$

Answer 1

$b_n \to l \to \ln b_n \to \ln l \to \dfrac{\ln b_1 + \ln b_2 +\cdots \ln b_n}{n} \to \ln l$ by Cesaro theorem, and you are done.