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.


Try to prove this:

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


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 $a\geqslant b$.

  • $\begingroup$ 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

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 $A_{n_0+k}^{1/(n_0+k}$.

  • $\begingroup$ 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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