# Convergence of Ratio Test implies Convergence of the Root Test

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...

• As I recall, you divide the $a_n$ into two parts: a final part in which the ratio is within $\epsilon$ of L and an initial part which, because its length is bounded, can be shown to not affect the result. There are a lot of limit-type results which are proved this way. – marty cohen Jan 27 '13 at 7:24
• @martycohen: Do you mean something like $|L-|a_{N+1}/a_N|\space |<\varepsilon$? I'm not sure I follow what you mean by dividing $a_n$ into two parts if that isn't what you mean. – Clayton Jan 27 '13 at 7:29
• Which question is this? Which number/chapter? – leo May 1 '13 at 2:15
• @leo: Chapter $1$, Exercise $17$. – Clayton May 1 '13 at 2:20
• – Martin Sleziak May 2 '14 at 15:15

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 $\lim_{n\to\infty}\sqrt[n]{|a_n|} \ge L.$

• That is excellent! Thanks! – Clayton Jan 27 '13 at 16:04

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 $$\lim_{n\to\infty} \sqrt[n]{a_n} = 0$$.

• @Clayton You don't have to, the $a_n$ in the statement of the theorem is $n$ in the proof. – Jakobian Jul 23 '19 at 19:19
• @Clayton Sorry, it should be ok now. I didn't realize I wrote 'exists' in the statement of Stolz theorem, it works even if the limit is infinite. – Jakobian Jul 23 '19 at 19:59
• Ah, very interesting! So you just need the limit to exist in the extended real numbers. $+1$ – Clayton Jul 23 '19 at 20:22