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...

  • $\begingroup$ 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. $\endgroup$ Jan 27, 2013 at 7:24
  • $\begingroup$ @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. $\endgroup$
    – Clayton
    Jan 27, 2013 at 7:29
  • $\begingroup$ Which question is this? Which number/chapter? $\endgroup$
    – leo
    May 1, 2013 at 2:15
  • 2
    $\begingroup$ @leo: Chapter $1$, Exercise $17$. $\endgroup$
    – Clayton
    May 1, 2013 at 2:20
  • 1
    $\begingroup$ See also: math.stackexchange.com/questions/69386/… $\endgroup$ May 2, 2014 at 15:15

3 Answers 3


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

  • 1
    $\begingroup$ That is excellent! Thanks! $\endgroup$
    – Clayton
    Jan 27, 2013 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 $\lim_{n\to\infty} \sqrt[n]{a_n} = 0$.

  • $\begingroup$ @Clayton You don't have to, the $a_n$ in the statement of the theorem is $n$ in the proof. $\endgroup$
    – Jakobian
    Jul 23, 2019 at 19:19
  • $\begingroup$ @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. $\endgroup$
    – Jakobian
    Jul 23, 2019 at 19:59
  • $\begingroup$ Ah, very interesting! So you just need the limit to exist in the extended real numbers. $+1$ $\endgroup$
    – Clayton
    Jul 23, 2019 at 20:22

Since $\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=L$, $\lim_{n\to\infty}\log\frac{|a_{n+1}|}{|a_n|}=\log L$.
So,for any positive real number $\epsilon$, there is a positive integer $N$ such that if $N\leq n$, then $\log L-\epsilon<\log\frac{|a_{n+1}|}{|a_n|}<\log L+\epsilon$.
Let $A:=\log\frac{|a_{2}|}{|a_1|}+\dots+\log\frac{|a_{N}|}{|a_{N-1}|}$.
If $N\leq n$, then $$A+(n-N+1)(\log L-\epsilon)<\log |a_{n+1}|-\log |a_1|<A+(n-N+1)(\log L+\epsilon).$$ Let $B:=A+\log |a_1|$.
If $N\leq n$, then $$B+(n-N+1)(\log L-\epsilon)<\log |a_{n+1}|<B+(n-N+1)(\log L+\epsilon).$$
If $N\leq n$, then
$$\frac{B}{n+1}+\frac{1+\frac{-N+1}{n}}{1+\frac{1}{n}}(\log L-\epsilon)<\log |a_{n+1}|^{\frac{1}{n+1}}<\frac{B}{n+1}+\frac{1+\frac{-N+1}{n}}{1+\frac{1}{n}}(\log L+\epsilon).$$
Since $\lim_{n\to\infty}\left(\frac{B}{n+1}+\frac{1+\frac{-N+1}{n}}{1+\frac{1}{n}}(\log L-\epsilon)\right)=\log L-\epsilon$ and $\lim_{n\to\infty}\left(\frac{B}{n+1}+\frac{1+\frac{-N+1}{n}}{1+\frac{1}{n}}(\log L+\epsilon)\right)=\log L+\epsilon$ and $\epsilon$ was any positive real number, $\lim_{n\to\infty}\log |a_{n+1}|^{\frac{1}{n+1}}$ exists and its value is $\log L$.


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