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I'd like your help with this:

The sequence $a_{_{n}}$ applies these condition:

$a_{_{n}}> 0$ for every $n \in \mathbb{N}$

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

I need to prove that $a_{n}$ is convergent, and it's limit is 0.

I tried to work with the fact that $a_{_{n}}> 0$ and (not successfully) show that $a_{n}> a_{n+1}$, and than to conclude what I need.

Thanks

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    Hint: Try to show that there exists $u>0$ and $N$ such that, for every $n\ge N$, $a_{n+1}\le(1-u)a_n$, and proceed from there. – Did Apr 25 '11 at 09:12
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    Just a comment to point out that if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ exists, then $\lim_{n\to\infty}\sqrt[n]{a_n}$ exists and $\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ (e.g., see http://math.stackexchange.com/questions/28476/finding-the-limit-of-n-sqrtnn/28487#28487). Therefore the result here follows from the result of your other recent question at http://math.stackexchange.com/questions/35007/lim-n-to-infty-sqrtna-n-1-a-n-geq-0-for-every-n-in-mathb. – Jonas Meyer Apr 25 '11 at 15:36
  • Possible duplicate of Showing that $a_n \to 0$ if $a_n/a_{n+1} \to l > 1$. – Jendrik Stelzner Sep 07 '18 at 10:47
  • @JendrikStelzner The question you linked to is more recent and doesn't really have better answers (they're all equivalent to the answer accepted below). Wouldn't it be better to close in the other direction? – Arnaud D. Sep 11 '18 at 12:39
  • @ArnaudD. I chose the linked question because I found it to have the best answer off all four, but this is probably just a matter of taste. The question can probably also be closed in another direction (although I don’t think that one has to necessarily chose the oldest one). – Jendrik Stelzner Sep 11 '18 at 13:12

2 Answers2

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Put $l:=\lim_{n\to+\infty}\frac{a_{n+1}}{a_n}$. We apply the definition of the limit with $\varepsilon :=\frac{1-l}2>0$. Hence we can find $n_0$ such that for $n\geq n_0$ $\frac{a_{n+1}}{a_n}\leq l+\frac{1-l}2 = \frac{l+1}2$. We get $0\leq a_{n+1}\leq a_n\frac{l+1}2$ hence $0\leq a_n\leq a_{n_0}\left(\frac{l+1}2\right)^{n-n_0}$ if $n\geq n_0$. Now you can conclude.

Davide Giraudo
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  • $0\leq a_n\leq a_{n_0}\left(\frac{l+1}2\right)^{n-n_0}$ is enough to show that $0\leq a_{n+1}\leq a_{n}$, why should I need the rest? to show that the lim is 0 with the squeez theorem? –  Apr 25 '11 at 10:05
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    Yes, the fact that the sequence is decreasing (and low bounded) shows that the limit exists, by it doesn't prove that this limit is $0$. – Davide Giraudo Apr 25 '11 at 10:13
  • Well, once you prove that the limit is finite, it can only be zero, otherwise $1= \frac{L}{L} = \lim_{n\to+\infty}\frac{a_{n+1}}{a_n} <1 $. – N. S. May 15 '11 at 18:44
  • @DavideGiraudo: Do you know what a compact element in a lattice is? – Minimus Heximus Sep 09 '14 at 19:08
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Edit:

Here is a partial answer which only proves the first half:

You can translate $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1$ into

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

But this means

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

for all $n > N$.

But $a_n > 0$ for all $n$ which means the sequence $(a_n)$ has a lower bound, therefore $(a_n)$ converges.

In the last step of reasoning I have used that a bounded monotonic sequence in $\mathbb{R}$ converges. For the statement and a proof of this look for example here.

The bounds of your sequence are $0$ (below) and $a_0$ (above).

Rudy the Reindeer
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    Sorry Matt but you made precisely the mistake to (not) make. First, your first displayed sentence is not how the hypothesis can be translated. Second, one cannot prove that the limit is zero from there. (For the correct proof see my comment on the post.) – Did Apr 25 '11 at 09:22
  • If you want to show that the limit is 0, you'll also need to use tha fact that limit of a_{n+1}/a_n is smaller than 1. (Not only the monotonicity of this sequence, which is a weaker property. There exist decreasing sequences such that this limit is equal to 1.) – Martin Sleziak Apr 25 '11 at 09:23
  • @Matt: Thank you, and will it be enough to write that decreasing and lower-bounded sequence converges to it's infimum? –  Apr 25 '11 at 09:23
  • @Didier, @Martin: I'm not proving that the limit is $0$! – Rudy the Reindeer Apr 25 '11 at 09:37
  • @Didier: sorry, can you explain a bit more detailed, I can't find my mistake, even reading your comment above, thank you! – Rudy the Reindeer Apr 25 '11 at 09:37
  • @Nir: No. But as for now you should use david's answer, it seems there might be something wrong in the reasoning with my answer. – Rudy the Reindeer Apr 25 '11 at 09:41
  • What you wrote is a translation of the hypothesis that $(a_n)$ is ultimately decreasing. The hypothesis the OP is interested in is translated in my comment to the main post. – Did Apr 25 '11 at 09:41
  • And @Martin's comment says it all as well. – Did Apr 25 '11 at 09:42
  • For a start you might ponder the difference between the assertions: (1) $x_n<1$ for every $n$, (2) there exists $u>0$ such that $x_n\le1-u$ for every $n$. – Did Apr 25 '11 at 09:44
  • @Didier: Ah! Yes but Martin's comment writes that I'm proving that the limit is zero, which I'm not. – Rudy the Reindeer Apr 25 '11 at 09:46
  • @Didier: I think I'll have to make my own, follow up question about this "translation". I still don't see what's wrong. – Rudy the Reindeer Apr 25 '11 at 09:49
  • Martin did not write what you say. – Did Apr 25 '11 at 09:51
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    You do see the difference between "$x_n<1$ for every $n$" and "$x_n\le1-u$ for every $n$", right? That is all there is to this, really. – Did Apr 25 '11 at 09:52
  • @Didier: yes, I see the difference. – Rudy the Reindeer Apr 25 '11 at 09:56
  • For what it's worth, what Matt wrote seems to prove a limit exists (though I wouldn't have used the word "translated" as he did - there is definitely some information is lost between the given information and the "translated" version). I don't see a mistake in his proof if this is his goal. On the other hand, I agree completely that the information "lost in translation" is crucial for actually showing the limit is 0. – Jason DeVito - on hiatus Apr 25 '11 at 15:32
  • @Jason: See the discussion on http://math.stackexchange.com/questions/35001. – Did Apr 26 '11 at 05:17
  • @Didier: I read the discussion, and I still agree with what I wrote above: Matt's answer proves a limit exists, but cannot be used to prove the limit is 0. If I am missig something, could you please be more specific? – Jason DeVito - on hiatus Apr 26 '11 at 05:51
  • @Jason The point of what you call the discussion is precisely that Matt's answer shows the limit exist and that the first step of the proof loses so much that it cannot be used to prove the limit is zero. – Did Apr 26 '11 at 06:19
  • @Didier: It seems as if we are on the same page after all :-). Thanks for clarifying! – Jason DeVito - on hiatus Apr 26 '11 at 07:45
  • @Jason: thank you for clarifying! – Rudy the Reindeer Apr 26 '11 at 11:05
  • Ah, so you finally did it! Better late than never... :) – t.b. May 15 '11 at 14:23
  • @Theo: actually I am doing it. True it is late but I've been awfully busy recently. Will post my own proof in my follow up on this. – Rudy the Reindeer May 15 '11 at 14:53
  • Very good! I'll have a look, then. – t.b. May 15 '11 at 14:55