Limits Application Question (Ratio Test Proof)

Question

I'm watching a video on ratio test proof and am confused about a statement.

For an infinite series

$$\sum_{n=1}^\infty a_n$$ $a_n > 0$, and $$L = \lim_{n\to \infty}\frac{a_{n+1}}{a_n}$$

If $L>1$, pick $\epsilon$, such that $L - \epsilon > 1$.

The video states that by definition of limit, there must be an $N$ such that for $n \geq N$, $a_{n+1}/a_n > L - \epsilon$. Why is this?

My process: We know L = lim as n -> infinity of (a sub [n+1])/(a sub n), so by definition of limit |(a sub [n+1])/(a sub n)-L| < L - epsilon, leading to 1)(a sub [n+1])/(a sub n)-L < epsilon or 2) (a sub [n+1])/(a sub n)-L > epsilon, but how do we know which of these is true?

Similarly, the video also says: If L<1, pick epsilon, such that L + epsilon < 1. And that by definition of limit, there must be an N such that for n>=N, (a sub [n+1])/(a sub n) < L + epsilon. For probably the same reason, why is this?

(Note: Please explain at my level, my level is multivariable calculus)

(video in question:https://youtu.be/2gi7pyQNxbM?si=3hvx0cegO5fO5q8D&t=257)

Answer 1

The video states that by definition of limit, there must be an $N$ such that for $n\geq N$, $\frac{a_{n+1}}{a_n} > L - \epsilon$. Why is this?

Because the definition of a limit is this:

Let $b_n$ be a sequence of real numbers. Then, we say that $$L=\lim_{n\to\infty} b_n$$ if and only if, for every $\epsilon > 0$, there exists some $N\in\mathbb N$ such that for all $n\in \mathbb N$, we know that if $n\geq N$, then $|b_n-L|<\epsilon$. Written with symbols, that means $$L=\lim_{n\to\infty} b_n\iff \forall\epsilon>0\exists N\in\mathbb N\forall n\in\mathbb N: n\geq N\implies |b_n-L|<\epsilon$$

Simply apply this definitio to $b_n=\frac{a_{n+1}}{a_n}$, and you get more or less the claim from the video. What you really get is that $|\frac{a_{n+1}}{a_n}-L|<\epsilon$, but that directly implies the claim in the video.

Answer 2

If $ \lvert a _ { n + 1 } / a _ n - L \rvert < \epsilon $, this means that $ a _ { n + 1 } / a _ n - L < \epsilon $ AND $ a _ { n + 1 } / a _ n - L > - \epsilon $. (For an absolute value to be close to zero, the number must not be too large and must not be too small.) So you don't have to decide which is true; the proof only needs one of these, but both are available. (And notice that it's $ - \epsilon $ in the second one.)

You can also write it as $$ - \epsilon < \frac { a _ { n + 1 } } { a _ n } - L < \epsilon \text , $$ and then the proof only needs the left-hand inequality (but both are true).

The second half is similar, except that this time, the proof only needs the right-hand inequality.