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This is a follow up to this.

Given $$0 \leq \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1$$

I think this is equivalent as saying

$$ \exists N: \forall n: n > N \implies \frac{a_{n+1}}{a_n} < 1$$

Apparently, this is wrong but I fail to see why. Can someone explain to me why? Thank you!

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It cannot be equivalent because $\exists N: \forall n>N \,\frac{a_{n+1}}{a_n}<1$ does't prove that the limit exists. – Davide Giraudo Apr 25 '11 at 10:09
@david: true but neither does the first line. – Rudy the Reindeer Apr 25 '11 at 10:11
Matt: In fact the present post is a follow up to a part of the page you link to, and furthermore, a part you chose to delete. So I am not sure the link explains anything at all (not to mention the fact that, as you know, I provided some explanations over there, which have now disappeared due to your deletion). – Did Apr 25 '11 at 10:28
Matt: To me this is as inept as can be (and all this reputation thing and the kinds of behaviour it generates is seriously beginning to annoy me...). In the case at hand you could simply add the mention Edit: The solution below is wrong because so and so at the beginning of your post, the content would not be lost and this would be (1) unambiguous (hence no more downvotes) and (2) pedagogical if you bother to expand so and so to explain where exactly you went wrong. – Did Apr 25 '11 at 10:56
This is to second @Didier's comment. But do follow his advice and expand on why exactly the argument is wrong (in your own words, at the moment it is buried in the comment thread) and try to say what you actually show. The point of this exercise would also be that you learn from that. In fact, I think one learns a whole lot in trying to identify the exact spots where one went wrong and like that one can try to avoid such mistakes in the future (recall the $\simeq$ vs. $=$ incident, I'm sure this won't happen to you again). Note that I'm not making fun of you, I'm very serious. – t.b. Apr 25 '11 at 11:31

If you want a counter-example, take $a_n:=\frac 1n$. We have $\frac{a_{n+1}}{a_n} = \frac n{n+1}< 1$ for all $n\geq 1$ but $\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = 1$.

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up vote 2 down vote accepted

I'm posting the proof of the claim in the original question in my own words:

claim: If $$ \forall n \in \mathbb{N}: a_n > 0$$ and $$ \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1$$


(i) $\{ a_n \}_n $ converges

(ii) $\lim a_n = 0$


$$ 0 \leq \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1 \implies \exists 0< s <1: \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = s$$

$$ \iff \forall \varepsilon > 0 \exists N: n > N \implies \frac{a_{n+1}}{a_n} \in [s-\varepsilon, s+\varepsilon]$$

Choose $\varepsilon$ s.t. $s+\varepsilon < 1$ and $s-\varepsilon > 0$. Then $$ 0 < s-\varepsilon \leq \frac{a_{n+1}}{a_n} \leq s+\varepsilon < 1$$

$$ \iff (s-\varepsilon) a_n \leq a_{n+1} \leq (s+\varepsilon) a_n$$

and setting $\delta := s+\varepsilon$:

$$ \implies a_{n+k} \leq \delta a_{n+k-1} \leq \delta^2 a_{n+k-2} \leq \dots (\forall n > N)$$

$$ \implies a_{n+k} \leq \delta^{k} a_n (\forall n > N)$$

Then $0 \leq a_{n+k} \leq a_n \delta^k$ and $a_n \lim_k \delta^k = 0$ implies

$$\lim_k a_{n+k} = 0$$

$$ \implies \lim_n a_n = 0$$

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This looks more or less okay (at least the right idea is there). However, I don't quite follow you after "Then..." (the equivalences are a bit obscure and probably wrong). You should check this again (and please tell us what $\delta$ is supposed to be!). Note also that the inequality $\frac{a_{n+1}}{a_{n}} \leq s + \varepsilon =: \lambda$ only holds for $n \gt N$. Thus, using your idea you get $a_{N+k} \leq \lambda a_{N+k-1} \leq \cdots \leq \lambda^{k-1} a_{N+1} \;\xrightarrow{k \to \infty}\;0$, as $\lambda \lt 1$. – t.b. May 15 '11 at 15:46
@Theo: many thanks for looking at my proof. I'm going to add some more detail above. – Rudy the Reindeer May 15 '11 at 16:31
This is still not quite correct. Look at my previous comment again! – t.b. May 15 '11 at 16:45
@Theo: errr...yes? I specified $\delta$ and I added $(\forall n > N)$ in the two lines after that. I think this is the same as writing $a_{N + k}$. – Rudy the Reindeer May 15 '11 at 17:07
The problem is that the inequality $a_{n+1} \leq \delta^{n-1} a_{0}$ is wrong for several reasons. First, the exponent of $\delta$ is wrong, but much more importantly you can only "go back" up to $a_{N+1}$, not all the way to $a_0$ since for $n \leq N$ the hypothesis $a_{n+1}/a_{n} \leq \delta$ is no longer satisfied in general. – t.b. May 15 '11 at 18:00

If you know that limits of sequences $(x_n)$, $(y_n)$ exist and $x_n<y_n$ for each $n$ (or for large enough $n$) you can only deduce that $$\lim_{n\to\infty} x_n \le \lim_{n\to\infty} y_n.$$ (You might try to prove this as an exercise, if you want.)

Simple counterexexample showing the strict inequality between limits need not be true: Put $x_n=1-\frac1n$ and $y_n=1$. Both sequences converge to $1$.

Your question is a special case of the application of this general principle, with $x_n=\frac{a_{n+1}}{a_n}$ and $y_n=1$.

This is also somewhat related to squeeze lemma (AKA two policemen and a drunk theorem).

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Thanks a lot Didier, I've corrected the answer. – Martin Sleziak Apr 25 '11 at 10:33
Thanks to you for the squeeze lemma link, now I know how to translate Théorème des gendarmes in English... – Did Apr 25 '11 at 10:46

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