TITLE: Convergence of Ratio Test implies Convergence of the Root Test
QUESTION [35 upvotes]: In Elias Stein and Rami Shakarchi's Complex Analysis textbook, we have the following exercise:

Show that if $\{a_n\}_{n=0}^\infty$ is a sequence of complex numbers such that $$\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=L,$$ then $$\lim_{n\to\infty}|a_n|^{1/n}=L.$$ 

I've been trying to prove this with no luck. The only thing I've thought of doing is $$\lim_{n\to\infty}\left(\frac{|a_{n+1}|^n}{|a_n|^n}\right)^{1/n},$$but this hasn't lead me anywhere except dead ends. Will someone provide a hint for me about how to proceed? Thanks!
Minor update: I don't know if it's helpful yet, but I know we can write the limit as $$\lim_{n\to\infty}\left(\frac{|a_{n+1}a_n\cdots a_0|}{|a_n\cdots a_0|}\cdot\frac{1}{|a_n|}\right).$$This reminds me a lot of the geometric mean, which even has the exponents I'm trying to get...

REPLY [4 votes]: This can also be proven by using Stolz theorem as shown in Fichtenholz's 'Differential and Integral Calculus'. We also need to know some facts about logarithms and exponentiation.

Stolz Theorem: Suppose that $(a_n)_{n\geq 1}$ and $(b_n)_{n\geq 1}$ are sequences of real numbers. Assume that $(a_n)_{n\geq 1}$ is a strictly increasing sequence, divergent to $\infty$, and that $$\lim_{n\to\infty} \frac{b_{n+1}-b_n}{a_{n+1}-a_n}\ \text{exists (possibly is infinite)}. $$ Then $\lim\limits_{n\to\infty} \frac{b_n}{a_n}$ exists (which might also be infinite) and $$\lim_{n\to\infty} \frac{b_{n+1}-b_n}{a_{n+1}-a_n} = \lim_{n\to\infty} \frac{b_n}{a_n}. $$

Let's assume that $(a_n)_{n\geq 1}$ is a sequence of positive numbers and $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}$ exists. Then using Stolz theorem:
$$\lim_{n\to\infty}\ln(a_n^{1/n}) = \lim_{n\to\infty}\frac{\ln(a_n)}{n} = \lim_{n\to\infty} \big(\ln(a_{n+1})-\ln(a_n)\big) = \lim_{n\to\infty} \ln\left(\frac{a_{n+1}}{a_n}\right),$$
from which immediately $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \sqrt[n]{a_n}. $$
Note: For the case when $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 0$ the proof is still valid, we still get that $\lim_{n\to\infty} \ln(\sqrt[n]{a_n}) = -\infty$, which can happen if and only if $\lim_{n\to\infty} \sqrt[n]{a_n} = 0$.