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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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Rudy the Reindeer
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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
  • I meant the limit $\lim_{n\to+\infty}\frac{a_{n+1}}{a_n}$ (it was ambiguous). – Davide Giraudo Apr 25 '11 at 10:15
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    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
  • And @david's example is perfect. – Did Apr 25 '11 at 10:29
  • @Didier: yes, I know. I would've left it but I don't want to accumulate down votes : / I'm sorry, I would've kept the comments visible if it had been possible. – Rudy the Reindeer Apr 25 '11 at 10:48
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    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
  • @Didier: I think you are right, especially about behaviour generated by the rating system. For exactly this reason I don't down vote. I've undeleted my answer in the other thread. – Rudy the Reindeer Apr 25 '11 at 11:06
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    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
  • Matt: OK. Do not forget the Here what is wrong with the solution below part--for the reasons Theo explained. (And I am not making fun of anybody here either. At all.) – Did Apr 25 '11 at 11:50

3 Answers3

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

Davide Giraudo
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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 $$ \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1$$

then

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

(ii) $\lim a_n = 0$

proof:

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

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

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

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

$$ \implies \lim_n a_n = 0$$

Rudy the Reindeer
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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
  • @Theo: true! I didn't notice. But I think it doesn't make any difference: it'll just be a different finite $a_i$ of the sequence so the limit of $\delta^k$ times $a_i$ will still be zero. I want to correct it anyway though. – Rudy the Reindeer May 15 '11 at 19:34
  • Right, it doesn't make much difference, but still it's wrong... While you're at it: Don't write $\lim_{n \to \infty} a_{n}$ in the penultimate line since this implicitly assumes that the limit exists. Show that $0 \leq a_{N+k} \leq \delta^{k-1} a_{N+1}; \xrightarrow{k \to \infty} ; 0$ and hence the limit of $a_{n}$ exists and is equal to zero by sandwiching or squeezing or whatever you call that result. – t.b. May 15 '11 at 19:40
  • Okay, we're almost there now! I'd prefer fixing $N$ and writing $a_{N+k} \leq \delta^{k} a_{N}$. Then as a last point, take into account my last remark and I'll let you off the hook. – t.b. May 15 '11 at 19:44
  • But Theo, is that not what I'm doing already? Exactly on the penultimate line and then on the last one I conclude with $\lim_n a_n = 0$. – Rudy the Reindeer May 15 '11 at 19:45
  • As I said, as long as you don't know that the limit exists, the expression $\lim_{k \to \infty} a_{n+k}$ does not make sense. Look at my penultimate comment and note the difference. – t.b. May 15 '11 at 19:47
  • Ah, you mean you're saying the sequence is eventually monotonically decreasing and bounded below by zero, and hence its limit exists? Then you should absolutely say that! You know, it is clear to me, but it is not clear to me that it is clear to you, that's why I'm so insistent. – t.b. May 15 '11 at 19:51
  • I could please you and fix $N$ and write $a_{N+k}$ but I want to use my own words and I think it's correct for any $n>N$ so I'm gonna leave it as is. – Rudy the Reindeer May 15 '11 at 19:56
  • I had proved that the limit exists in my answer to the original post but at the moment I'm trying to do it assuming that I don't know that already so you may insist. : ) – Rudy the Reindeer May 15 '11 at 19:57
  • first comment: okay, I can live with that. second comment: I'm just looking at this post... :) – t.b. May 15 '11 at 20:00
  • Very good. I think I'm happy now... – t.b. May 15 '11 at 20:31
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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). http://en.wikipedia.org/wiki/Squeeze_theorem

Martin Sleziak
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