# $\lim_{n\to \infty }\frac{a_{n+1}}{a_{n}}< 1$, $a_{_{n}}> 0$- does $a_{_{n}}$ converge?

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 9:12
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 math.stackexchange.com/questions/28476/…). Therefore the result here follows from the result of your other recent question at math.stackexchange.com/questions/35007/…. –  Jonas Meyer Apr 25 '11 at 15:36

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.

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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? –  user6163 Apr 25 '11 at 10:05
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? –  user795571 Sep 9 '14 at 19:08

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).

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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 9:22
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 9:23
@Matt: Thank you, and will it be enough to write that decreasing and lower-bounded sequence converges to it's infimum? –  user6163 Apr 25 '11 at 9:23
@Didier, @Martin: I'm not proving that the limit is $0$! –  Rudy the Reindeer Apr 25 '11 at 9:37
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 9:52