Proof that if $\lim_{n \to \infty } \frac{a(n+1)}{a(n)} =0$ for $a:\mathbb{N}\mapsto K, a(n)>0 \forall n\in \mathbb{N}$, then $\lim(a)=0$

Question

I want to prove:

Is $K$ an ordered field and $a:\mathbb{N}\rightarrow K$ with $a(n)>0\;\forall n\in \mathbb{N}$. If $\lim_{n \to \infty} \frac{a(n+1)}{a(n)} =0$, then $\lim a=0$.

Please let me know what you think about my solution:

Since I don't want to use that $K$ is archimedean, I can't go the usual way by picking $\epsilon=1/2$, using an induction proof and the sandwich theorem.

So instead of proving "A->B" I show that "not B -> not A".

So let $\lim a =b$ with $b\neq0$. Then $\lim_{n \to \infty} \frac{a(n+1)}{a(n)}=\frac{\lim_{n\to\infty} a(n+1)}{\lim_{n\to\infty} a(n)} = \frac{b}{b}=1 \neq 0$. qed.

Thank you!

Answer 1

A sketch may be as follows:

Assuming $a_n\geq0$ for all $n$. For all n large enough, say $n\geq N$

$$ a_{n+1}\leq \frac12 a_n $$ Then $$ a_{N+m}\leq 2^{-m}a_{N} $$

The rest should be easy.

Answer 2

What you need to prove here is that for any given element $\epsilon$ of $K$, there is a positive integer $N$ such that $n>N$ implies $a_n < \epsilon$.

Since $\lim_{n \rightarrow \infty} \frac {a_n+1}{a_n} = 0$, there must be some integer M, such that for all $n>M$, $\frac {a_n+1}{a_n}<1$.Then we can prove inductively that for all $n>M$, $a_n<a_M$. Now again because $\lim_{n \rightarrow \infty} \frac {a_n+1}{a_n} = 0$, then there must exist some integer $N_1$ such that $n>N_1$ implies $\frac {a_n+1}{a_n}<\frac \epsilon {a_M}$.

Now define $N= \min (M+2,N_1)

Since $a_{N-1}<a_M$, we have that $$a_N = a_{N-1}\frac {a_N}{a_{N-1}}<a_M \frac {a_N}{a_{N-1}}<a_M\frac {\epsilon}{a_M}=\epsilon$$.