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...
By definition of limit, for each $\varepsilon>0$ there exists $N$ s.t. $$n>N \implies \left| \left| \frac{a_{n+1}}{a_n} \right|-L \right|<\varepsilon.$$ So $$|a_n|=\frac{|a_n|}{|a_{n-1}|}\cdots \frac{|a_{N+1}|}{|a_N|} |a_N|<(L+\varepsilon) ^{n-N} |a_N|$$ Take the $n$th root of both sides of the inequality. Then we get $$\sqrt[n]{|a_n|} <(L+\varepsilon)^{1-N/n}\sqrt[n]{|a_N|}.$$ Taking $n\to\infty$ then $$\lim_{n\to\infty}\sqrt[n]{|a_n|} \le L+\varepsilon.$$ Since $\varepsilon$ is arbitrary, we get $\lim_{n\to\infty}\sqrt[n]{|a_n|} \le L.$ Likewise we can get $\lim_{n\to\infty}\sqrt[n]{|a_n|} \ge L.$