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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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    @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).2018-09-11

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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.

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    @DavideGiraudo: Do you know what a compact element in a lattice is?2014-09-09
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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).

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    Very good! I'll have a look, then.2011-05-15