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I'm trying to solve the following problem and while looking around I found a couple of notes (Link) by Pete L. Clark where it's discussed but can't really comprehend his proof.

The problem: Prove that $ \lim\inf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} \leq \lim\inf\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \lim\sup\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$

His proof, as I understood it, is as follows: Let $r > \lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} $, so there's an $n_0$ so that for all $k \geq 1$ then $ \frac{A_{n_0+k}}{A_{n_0 + k - 1}} < r $.

From that we get $ A_{n_0 + k} < r A_{n_0+k-1} $ and that implies $A_{n_0 + k} < r^{k} A_{n_0} $. Rewriting that as $A_{n_0+k} ^{\frac{1}{n_0+k}} < r (\frac{A_{n_0}}{r^n})^{\frac{1}{n_0+k}}$ and by letting $ k \rightarrow \infty$ we see that $\lim\sup\limits_{k\rightarrow \infty} A_{n_0+k} ^{\frac{1}{n_0+k}}$ is at most $r$.

So, we have that $r > \lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} $ and $ r \geq \lim\sup\limits_{n\rightarrow \infty} (A_n)^{1/n}$

He says that that implies $\lim\sup\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$, but I really don't see how that's the case to be honest. And I'm also not too sure how the analogous argument for $\lim\inf$ would look like.

I'm sure the proof is correct, as I've seen a lot of people recommend that particular set of notes, but I just can't seem to grasp it and it looks so simple.

Any help, both with this particular proof and with any other that applies, would be greatly appreciated.

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Try to prove this:

Suppose that for each $r>b$, we have $a\leqslant r$. Then $a\leqslant b$.

Equivalently

Suppose that for each $\epsilon >0$, we have $y\leqslant \epsilon$. Then $y \leqslant 0$.

Pete (who is a user here!) is using this with $b=\limsup\limits_{n\to\infty} \dfrac{ A_{n+1}}{A_n}$ and $a=\limsup\limits_{n\to\infty} A_n^{1/n}$.

For the argument using $\liminf$; you start with $\liminf\limits_{n\to\infty}\frac{A_{n+1}}{A_n}$ and choose $r$ smaller than this. The steps are the same but with signs reversed. Then you use

Suppose that for each $r<b$, we have $a\geqslant r$. Then $a\geqslant b$.

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  • $\begingroup$ Thanks! I confess it took me a while longer than what it should have, for some reason. $\endgroup$ – Bananas Aug 21 '13 at 6:32
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The idea is very simple, though understandably, it tends to drown in other detail: You are given numbers $U$ and $V$ so that whenever $r>V$, it is also true that $r>U$. Then $U\le V$. For if $U>V$ then you can pick $r$ with $U>r>V$ to get a contradiction.

In this case, $U=\limsup\limits_{n\to\infty}(A_n)^{1/n}$ and $V=\limsup\limits_{n\to\infty}(A_{n+1}/A_n$.

This sort of technique is very useful because working with strict inequalities gives you a bit of wiggle room. In this case, the wiggle room is used to accomodate the term $A_{n_0+k}^{1/(n_0+k}$.

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  • $\begingroup$ Thanks. For some reason I didn't though that was true. That's what I get for doing stuff at 3am. $\endgroup$ – Bananas Aug 21 '13 at 6:34

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