# Proof attempt to the ratio test for sequences

I'm trying to prove the ratio test for sequences. Here's what I got:

If $\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} = L < 1$ and $a_n>0 \;\ \forall n$ then $a_n$ is bounded below by $0$. Also there's $N$ so that forall $n>N$, $a_{n+1}<a_n$. Therefore, the sequence is decreasing and bounded below so it must converge.

Now, according to the test, $\lim \limits_{n \to \infty} a_n = 0$. Why is that?

If we note that the limit statement is formally that for any $\epsilon>0$ we may select $N$ so that for $n>N$ we have

$$\left|{a_{n+1}\over a_n}-L\right|<\epsilon$$

but then we see, by multiplying both sides by $a_n>0$, that

$$\left|a_{n+1}-La_n\right|<\epsilon a_n$$

then we have $a_{n+1}-La_n<\epsilon a_n\implies a_{n+1}<(L+\epsilon)a_n$

choose $0<\epsilon <(1-L)$--this is positive since $0<L<1$. Then $L+\epsilon <1$ and we see then that

$$0<a_{n+m}<(L+\epsilon)^ma_n$$

and as $m\to\infty$ we have $a_{n+m}\to 0$ by the squeeze theorem.