Why does the Ratio Test prove that this particular sequence converges on 0?

Question

I was recently reading How to Think about Analysis by Lara Alcock in preparation for taking Real Analysis in the winter, and ran into the following theorem regarding sequences that either tend towards infinity or converge on a particular number, specifically the Ratio Test.

Suppose that $(a_n)$ is a sequence such that $(a_{n+1}/a_n) \rightarrow l$. Then:

1. If $-1 < l < 1$ then $(a_n) \rightarrow 0$.
2. If $l > 1$ and $a_n > 0 \forall n \in \mathbb{N}$ then $(a_n) \rightarrow \infty$.
3. If $l > 1$ and $a_n < 0 \forall n \in \mathbb{N}$ then $(a_n) \rightarrow -\infty$
4. If $l < -1$ then the sequence neither converges or tends to $\pm \infty$.
5. If $l = 1$ we get no information.

Two specific examples she gives of sequences that converge to $0$ are $$\left({n^2 \over 2^n}\right)$$ and $$\left({6^n \over n!}\right)$$

For the latter sequence, $$\left({6^n \over n!}\right) = {6 \over 1}, {36 \over 2}, {216 \over 6}, {1296 \over 24}, {7776 \over 120},\ldots$$

I can see for the ratios I test that that the ratios tend to get smaller over time, which isn't too surprising. For example, $${36 \over 2} \div {6 \over 1} = 3, {216 \over 6} \div {36 \over 2} = 2, {1296 \over 24} \div {216 \over 6} = 1.5$$

The author also gives the intuition of why this sequences approaches $0$ by pointing out that $${6 \times 6 \times\cdots \times 6} \over {1 \times 2 \times \cdots \times n}$$

This makes it intuitively obvious why this denominator grows much more quickly than the numerator, which makes it seem a lot more sensible why this should approach $0.$

So, I understand that the ratio between each two consecutive terms is decreasing (if I'm reading this correctly), which isn't too surprising, and I understand the intuition behind why this sequence approaches $0.$ However, I'm a little confused as to how this actually "meets" the first test given that I gave several examples where $(a_{n+1}/a_n) > 1$ - for example, ${36 \over 2} \div {6 \over 1} = 3$, which is obviously greater than $1.$ Naturally, eventually the ratio will be less than $1$, but doesn't the theorem kind of imply that the ratio of any two consecutive elements of the sequence will have a ratio greater than $-1$ and less than $1$?

Also, how does this account for sequences that converge on things other than $0$, $- \infty$, or $\infty$?

Answer 1

To say that a sequence $\{b_n\}_{n \in \mathbb{N}}$ converges to a limit $L$ is to say that, for all $\varepsilon > 0$, there is an $N \in \mathbb{N}$ such that $n > N$ implies $|b_n - L| < \varepsilon$.

Rephrased into colloquial English: For any positive number $\varepsilon$, the sequence of those $b$-subscript elements eventually gets within $\varepsilon$ of $L$, and stays at least that close to the limit. The key word in this description is the same term that you italicized:

Naturally, eventually the ratio will be less than $1$, but doesn't the theorem kind of imply that the ratio of any two consecutive elements of the sequence will have a ratio greater than $−1$ and less than $1$?

No, the theorem does not say this. The theorem's hypothesis is precisely that the sequence made from a ratio of consecutive terms converges to $l$, which means that eventually this sequence (i.e., these ratios) will be within $\varepsilon$ of $l$.