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If $\{a_{n}\}_{n\geq 1}$ is a decreasing sequence of real numbers, $a_{n}\in (0,1)$ and $\lim_{n\to \infty} a_{n}=0$. What we can say about $$\lim_{n\to\infty}\frac{a_{n-1}}{a_{n}}$$

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If it exists, it is greater than $1$. Don't think you can say much more than that. – Thomas Andrews Jul 25 '12 at 18:21
(That should be "greater than or equal to $1$, of course, as I showed below.) – Thomas Andrews Jul 25 '12 at 18:49
up vote 8 down vote accepted

Probably not much.

Let $a_n = b^{-n}$ for any constant $b > 1$, so then $a_n \in (0, 1)$. Then,

$$\lim_{n \to \infty} \frac{a_{n-1}}{a_n} = \lim_{n \to \infty} \frac{b^{-n + 1}}{b^{-n}} = \lim_{n \to \infty} b = b.$$

So, any real number greater than 1 is possible. And, if $a_n = \frac{1}{n}$, then a limit of 1 is also possible. And, using $a_n = b^{-n^2}$, we see $\infty$ is also a possible limit. The limit would have to be at least 1, in general, if $a_n$ is decreasing.

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Let $b_1,...,b_n,...$ be any sequence of positive real numbers such that $\sum_{i=1}^\infty b_i$ diverges.

Then let $a_0 = 1$ and $a_{n+1} = \frac{a_n}{1+b_n}$.

Then $a_{n-1}/a_{n}= 1+b_{n-1}$.

Now, since $\sum b_i$ diverges, $\prod_{i=1}^\infty (1+b_i)$ diverges.

But $a_{n} = \frac{1}{\prod_{i=1}^n (1+b_i)}$

So $a_{n}\to 0$.

On the other hand, $\lim a_{n-1}/a_n = \lim_{n\to\infty} (1+b_n)$

Clearly, that can converge, diverge, bop around. It can even converge to $1$, by setting $b_n=\frac{1}{n}$.

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On the second-to-last line, wouldn't it be $\lim_{n\to\infty} \frac{1}{1+b_n}$? – Eugene Shvarts Jul 25 '12 at 20:12
Ah, I got the earlier line backwards. I fixed it. It should have been $a_{n-1}/a_n = 1+b_{n-1}$, and I had the numerator and denominator switched – Thomas Andrews Jul 25 '12 at 20:16

We show that the behaviour of the $\frac{a_{n-1}}{a_n}$ can be extremely chaotic.

In particular, we show that there is a decreasing sequence of positive reals, with limit $0$, such that every real number in the interval $[1,\infty)$ is a limit of a subsequence of the $\frac{a_{n-1}}{a_n}$.

Let $r_0,r_1,r_2,\dots$ be an enumeration of the positive rationals, and let $s_n=\sum_{i=0}^n r_i$. Then $(s_n)$ is an increasing positive sequence, with limit $\infty$.

Let $a_n=e^{-s_n}$. Then $\frac{a_{n-1}}{a_n}=e^{r_n}$.

For any real number $x\ge 1$, there is a subsequence of $(r_n)$ which has limit $\log(x)$, and the result follows.

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