26
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – marty cohen Jan 27 '13 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 '13 at 7:29
  • $\begingroup$ Which question is this? Which number/chapter? $\endgroup$ – leo May 1 '13 at 2:15
  • 2
    $\begingroup$ @leo: Chapter $1$, Exercise $17$. $\endgroup$ – Clayton May 1 '13 at 2:20
  • 1
    $\begingroup$ See also: math.stackexchange.com/questions/69386/… $\endgroup$ – Martin Sleziak May 2 '14 at 15:15
34
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ That is excellent! Thanks! $\endgroup$ – Clayton Jan 27 '13 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.