0

Let $(c_n)_{n=m}^{\infty}$ be a sequence of positive numbers. I am trying to prove that if

$ \liminf_{n \to \infty} c_{n+1}/c_n = \infty $, then
$\liminf_{n \to \infty} {c_n}^{1/n} = \infty $.

Context: this is for Exercise 7.5.1 in Tao's Analysis I, where we have to prove the inequality $\liminf_{n \to \infty} c_{n+1}/c_n \leq \liminf_{n \to \infty} {c_n}^{1/n}$,

for $(c_n)_{n=m}^{\infty}$ a sequence of positive numbers. I was able to prove the inequality when $\liminf_{n \to \infty} c_{n+1}/c_n$ is finite (by adapting a similar proof that Tao gives in the same chapter), but I'm stuck on the infite case.

Any hints? Thank you.

Jonas
  • 107
  • 1
    What did you do for the finite case? – Keen-ameteur Apr 18 '23 at 14:11
  • Hint: $\liminf c_{n+1}/c_n =+\infty\implies\lim c_{n+1}/c_n =+\infty\implies\lim\sqrt[n]{c_n}=+\infty.$ – Anne Bauval Apr 18 '23 at 14:45
  • I believe it boils down to $\frac{c_{n+1}}{c_n}\gt \frac {c^{\frac{n+1}n}_n}{c_n}\forall n\ge N$. – WindSoul Apr 18 '23 at 15:14
  • This is special case of https://math.stackexchange.com/q/69386/42969 – Martin R Apr 18 '23 at 19:10
  • @WindSoul I do not understand your comment. – Anne Bauval Apr 19 '23 at 09:57
  • I shall vote to close this post as a duplicate of the one indicated by @MartinR : this answer proves $\limsup a_n^{1/n} \le \limsup\frac{a_{n+1}}{a_n}$ and hints that $\liminf\frac{c_{n+1}}{c_n} \le \liminf c_n^{1/n}$ is an immediate consequence (take $c_n=\frac1{a_n}$). This solves your problem, since $\liminf\frac{c_{n+1}}{c_n}=\infty\iff \limsup\frac{a_{n+1}}{a_n}=0$ and $\limsup a_n^{1/n}=0\iff \liminf c_n^{1/n}=\infty.$ I think we should both delete our answers. I shall do so as soon as Jonas acknowledges my last comment below my answer. – Anne Bauval Apr 19 '23 at 10:06
  • 1
    @AnneBauval I find the proof suggested by Oliver to be more intuitive and satisfying. But yes, the proof in the linked answer works as well. So I'd be fine with marking this post as a duplicate, if this is in accordance with moderation policy here. – Jonas Apr 19 '23 at 10:59
  • Please does anyone understand @WindSoul 's comment above? – Anne Bauval Apr 20 '23 at 06:00

1 Answers1

1

If $\liminf_n\frac{c_{n+1}}{c_n}=\infty$, then for any $b>0$, there is $N$ such that for $n\geq N$, $\frac{c_{n+1}}{c_n}>b$ (why?). Hence $$\sqrt[n]{c_n}>bb^{-N/n}\sqrt[n]{c_N}\qquad\text{for all}\quad n> N$$ this implies that $\liminf_n\sqrt[n]{c_n}\geq b$. This holds for any $b>0$.

Mittens
  • 39,145
  • Thank you for your reply! Unfortunately I don't fully follow. I can justify everything except for the last step. I believe there is a little error, since for $n = N$, ${c_n}^{1/n} > bb^{-N/n} {c_N}^{1/n}$ is false. But this can of course be easily fixed by restricting to $n \geq N + 1$.

    Anyway, about the last step: I understand that we'll get the result if $b^{N/n}{c_N}^{1/n} \geq 1$ for all $n \geq N + 1$, but I don't see how that follows from the previous line. Could you maybe elaborate?

     – Jonas Apr 18 '23 at 19:00
  • @JonasE.: From $c_{m+N}>b c_{m+N-1}>b^2 c_{m+N-2}>\ldots>b^m c_N$. Set $n=m+N$. Observe that $b^{M/n}\xrightarrow{n\rightarrow\infty}1$ and $c^{1/n}_N\xrightarrow{n\rightarrow\infty}1$. Thus the conclusion $\liminf_n\sqrt[n]{c_n}\geq b$. – Mittens Apr 18 '23 at 21:15
  • After lots of fiddling around I finally managed to grind out a proof. It turned out to be equivalent to yours, so I guess I unconsciously followed your reasoning after all. Thank you very much for your help! – Jonas Apr 18 '23 at 21:32
  • @JonasE.: glad you worked it out by yourself. – Mittens Apr 18 '23 at 22:12