Confusion about $\limsup$ properties in proof of ratio test

Question

In the proof of the ratio test in my notes, which is similar to the one here, the first step revolves around an obvious statement which is not proved. However, for whatever reason, I can't make out why it is at all obvious!

Note for our case, $a_n\geq0$ for all $n$. The comment in question is that $$\limsup_{n\to\infty} \frac{a_{n+1}}{a_n}<1$$ implies that there exists $\epsilon>0$ such that $$\frac{a_{n+1}}{a_n}<1-\epsilon$$ for all $n\geq N$, for some $N$.

I have also seen this post which seems to have include the same comment, but it doesn't seem to explain it in detail.

I can't figure out why this is true, but I suspect it is obvious, so any guidance would be much appreciated!

Edit: Our definition of $\limsup$ is $\limsup_{n\to\infty} a_n = \lim_{N\to\infty}\sup_{n\geq N} a_n$, or that it is the maximal limit point of $a_n$.

Answer 1

This can be argued very closely from definitions, although it's just making rigorous the intuitive notion that the ratios must maintain some distance from $1$, as otherwise they would tend to at least $1$, making their $\limsup$ at least $1$.

Let $\ell = \limsup_{n \to \infty} a_{n + 1}/a_n$. We are given that $\ell < 1$, so certainly $\tfrac 12(1 - \ell) > 0$.

Since $\ell = \lim_{m \to \infty} \sup_{n \ge m} a_{n + 1}/a_n$, by the definition of the limit of a sequence (letting $\varepsilon = \tfrac 12(1 - \ell)$), we can find some $N$ such that for all $m \ge N$, $\sup_{n \ge m} a_{n + 1}/a_n - \ell < \tfrac 12(1 - \ell)$. (Here I have omitted the modulus as the sequence $(\sup_{n \ge m} a_{n + 1}/a_n)$ is decreasing as a sequence in $m$, so $\ell \le \sup_{n \ge m} a_{n + 1}/a_n$).

Particularly, taking $m = N$, we get $\sup_{n \ge N} a_{n + 1}/a_n - \ell < \tfrac 12(1 - \ell)$.

But then, we have $\sup_{n \ge N} a_{n + 1}/a_n < \tfrac 12(1 + \ell) = 1 - \tfrac 12(1 - \ell)$. Since $\sup$ is an upper bound, for any $k \ge N$, we get $a_{k + 1}/a_k \le \sup_{n \ge N} a_{n + 1}/a_n < 1 - \tfrac 12(1 - \ell)$.

So $N$ is our $N$, and our $\varepsilon$ is $\tfrac 12(1 - \ell)$.

You might notice that this is actually a general fact about $\limsup$ of a sequence with a strict upper bound. We didn't need to use positivity of $a_n$ for this step of the proof.

Answer 2

let us call the $limsup$ as $p$, now $p<1$ , define $e =(1-p)/2$ so $ 1>1 - e>p$ , if all point after some certain $n$ didn't lie below $1-e$, (let us call $1-e$ as $\beta$), we can find infinitely many point of the seq which lies above $\beta$ (why?). we can always create a subseq which converges to a number (in the extended reals) $\alpha$ greater than or equal $\beta$ .So it is a limit of a subseq greater than the limsup, which is a contradiction!