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The ratio test is generally proved by letting,

$$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=L$$

And introducing a number $r$ where $L<r<1$.

Then saying that, there exists some integer $N$ such that for all $n \geq N$ we have,

$$\left|\frac{a_{n+1}}{a_n}\right|<r$$

Or in words, the ratio $\left|\frac{a_{n+1}}{a_{n}}\right|$ is eventually less $r$. The fact seems obvious by what limit means, and I don't see it proved anywhere I look. So I seek to prove it, here is my attempt:

Let $L<r<1$ so that $r-L>0$.

By definition of a limit to infinity,

For every $\epsilon>0$ there exists $N$ such for every that $n \geq N$ we have,

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

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

In particular take $\epsilon=r-L$ this gives,

$$L-r<\left|\frac{a_{n+1}}{a_{n}}\right|-L<r-L$$

Adding $L$ to both sides gives,

$$2L-r<\left|\frac{a_{n+1}}{a_{n}}\right|<r$$

Question: Is the proof valid? Is there something I missed that would make the proof more simple/ trivial?

I haven't seen the proof anywhere, in any textbooks, so I wonder if I'm just overthinking things.

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    $\begingroup$ It's a quite valid proof. $\endgroup$ – Bernard Apr 2 '17 at 22:48
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The proof is ok, here is an alternative wording, which I think feels more "constructive".

Since $L<1$ let's have $r=\frac{1+L}2$ then $L<r<1$.

For $\varepsilon = r-L>0$ there exists $N$ such that $\forall n>N,\ \bigg||\frac{a_{n+1}}{a_n}|-L\bigg|<\varepsilon\implies|\frac{a_{n+1}}{a_n}|<L+\varepsilon=r$

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