# Limits Application Question (Ratio Test Proof)

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)

– 5xum
Sep 25 at 8:11

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.

• I understand the statement |b_n-L|<epsilon from the definition of limits, but how do we get rid of the absolute value and choose one of the two possibilities? either b_n-L<epsilon or b_n-L>epsilon Sep 25 at 8:21
• @Pheonix $|b_n-L|<\epsilon$ is equivalent to $-\epsilon<b_n-L\land b_n-L<\epsilon$.
– 5xum
Sep 25 at 8:40

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.

• Thank you, I thought it was an OR instead of an AND. Sep 25 at 8:28
• If you have $|u|>a$, then you get an OR: $u>a$ or $u<-a$. Even for $|u|=a$, you mostly get an OR: $u=a$ or $u=-a$, and $a\geq0$; but the $a\geq0$ part is usually obvious. But for $|u|<a$, you get an AND: $-a<u<a$. (And the same rules work for weak inequalities $\leq,\geq$ too.) Sep 25 at 13:19