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?
$\endgroup$1 Answer
$\begingroup$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.
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