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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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    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.)2011-04-25

3 Answers 3

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If you know that limits of sequences $(x_n)$, $(y_n)$ exist and $x_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

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    Thanks to you for the squeeze lemma link, now I know how to translate *Théorème des gendarmes* in English...2011-04-25
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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 $\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = 1$.

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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$

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    Very good. I think I'm happy now...2011-05-15