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 $a_n$.

  • $\begingroup$ What is the definition of $\limsup$? $\endgroup$ – Peanut Jun 27 '20 at 12:24
  • $\begingroup$ @Dude Clarified in edit $\endgroup$ – Benjamin Jun 27 '20 at 12:28
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    $\begingroup$ Yes. So assume on the contrary that your inequality does not hold. Is it true that $\lim sup \frac{a_{n+1}}{a_n} < 1$? No because that means that an infinite number of those ratios is getting close to something greater than or equal to $1$. $\endgroup$ – Peanut Jun 27 '20 at 12:39

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.

  • $\begingroup$ Can one think of it this way? $\sup a_{n+1}/a_n < 1+\epsilon$ for all $\epsilon>0$, so we know that $\sup a_{n+1}/a_n < 1$ and hence $a_{n+1}/a_n < 1$, so this ratio must be less than or equal to $1-\epsilon$, for some $\epsilon>0$. $\endgroup$ – Benjamin Jun 27 '20 at 14:31
  • $\begingroup$ @Benjamin, I don't follow exactly what you're saying. Do you mean $\limsup$? I don't see what the $1 + \varepsilon$ is doing. $x < 1 + \varepsilon$ for all $\varepsilon > 0$ doesn't imply that $x < 1$. It also seems like at the end you show that $a_{n + 1} / a_n < 1 - \varepsilon$ for some $\varepsilon$, but in your question you need to show that the same epsilon works for all $n$. Perhaps that's what you're arguing but it would be made clearer by saying something like "for all $n$". $\endgroup$ – Izaak van Dongen Jun 27 '20 at 14:40
  • $\begingroup$ You're right, of course, this was a silly argument. I follow your proof, but I don't see what's quite so 'obvious' about it. I'm trying to also understand why the statement is 'morally right', so to speak. $\endgroup$ – Benjamin Jun 27 '20 at 15:09
  • $\begingroup$ @Benjamin, The way I think of it is that the $\limsup$ of a sequence is a sort of "eventual upper bound". After a while, the sequence must quieten down and stay below any number above the $\limsup$. All this argument says is "if the eventual upper bound is strictly less than $1$, then we can pull the sequence down below $1$". Certainly if this claim were not true, then the sequence would be at least $1$ infinitely often, so it wouldn't make any sense for the eventual upper bound to be less than $1$. Maybe that helps. If not, don't worry. Analysis can be hard to build intuition about. $\endgroup$ – Izaak van Dongen Jun 27 '20 at 15:14
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    $\begingroup$ This does help, thank you. $\endgroup$ – Benjamin Jun 27 '20 at 15:18

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!


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