Follow up question about translation of a limit expression 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!
 A: 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 $\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = 1$.
A: 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
$$\lim_k a_{n+k} = 0$$
$$ \implies \lim_n a_n = 0$$
A: 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
