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I've been set the following question as part of my Real Analysis homework and I'm struggling with it:

Let $(a_n)_n\in\mathbb{N}$ be a sequence of positive numbers. Show that:

$$ \limsup_{n\rightarrow\infty} a_n^\frac{1}{n} \leq \limsup_{n\rightarrow\infty} (\frac{a_{n+1}}{a_n}) $$

I have reasoned that if the RHS takes value $\infty$ then the inequality is automatically satisfied so we can assume RHS $= \lambda < +\infty$

We got a hint in the lecture that went something like this: For $m \geq n$,

$a_m \leq a_n \lambda^{m-n} \Rightarrow a_m^{\frac{1}{n}} \leq (a_n \lambda^{m-n})^\frac{1}{n}$

And then taking supremums a and limsups but I'm confused at this point and I'm not really sure where to take t from here.

Any help would be greatly appreciated.

Surb
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Alex Modell
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2 Answers2

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Hint: Let $\varepsilon>0$. There exists an $N$ such that if $n\geq N$, you have $\frac{a_{n+1}}{a_n}\leq \lambda+\varepsilon$. Show that this imply that there exists $c$ independant of $n$ such that for $n$ large we have $a_n\leq c( \lambda+\varepsilon)^n$.

Kelenner
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You want to take everything to the power $\frac1m$, not $\frac1n$. Fix $\epsilon>0$. By assumption, there exists $n$ such that $\frac{a_{m+1}}{a_m}\le\lambda+\epsilon$ for all $m\ge n$. Inductively this gives us $$a_m\le(\lambda+\epsilon)^{m-n}a_n$$ for all $m\ge n$, and so $$a_m^{1/m}\le(\lambda+\epsilon)^{1-n/m}a_n^{1/m}.$$ Now you can consider $n$ fixed, so $a_n^{1/m}\to1$ and $(\lambda+\epsilon)^{-n/m}\to1$ as $m\to\infty$. Hence $$\limsup_{m\to\infty}a_m^{1/m}\le\lambda+\epsilon.$$ Since $\epsilon>0$ was arbitrary, the result follows.

Jason
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  • Hi Jason, I follow your argument right up until the last part. Could you please give some more detail on why the final stage implies the result? – Alex Modell Mar 17 '17 at 16:01
  • If $L\le\lambda+\epsilon$ for every $\epsilon>0$, then $L\le\inf_{\epsilon>0}(\lambda+\epsilon)=\lambda$. – Jason Mar 17 '17 at 16:09