Consider the sequence $a_n$. We say that $\lim a_n = c$ if for every $\epsilon > 0$, there exists a natural number $N$ such that, if $n \geq N$, then $|a_n - c| < \epsilon$. This definition suggests that $N$ will always be a function of $\epsilon$, including a constant function (in which case $N$ is independent of $\epsilon$). Therefore, I'm wondering if we can instead define the limit $\lim a_n = c$ as follows: We say that $\lim a_n = c$ if there exists a function $N : (0,\infty) \to \mathbb N$ such that if $|a_n - c| \geq \epsilon$, then $n < N(\epsilon)$. This definition does not seem that useful (and is likely missing some conditions), so I'm wondering if a better one can be constructed.
1 Answers
Yes, a function like this is called a modulus of convergence for the sequence. It's a theorem that every convergent sequence has a modulus of convergence (proof: let $ N ( \epsilon ) $ be the minimum value of $ N $ that works for $ \epsilon $ in the definition, using that every non-empty set of natural numbers has a minimum; or you could use the axiom of choice); and of course, any sequence with a modulus of convergence definitely converges. Any time you have a ‹for every $ \epsilon $, there exists› situation, you can replace this with a function of $ \epsilon $, so you can have moduli of continuity, moduli of uniform convergence, etc.
Most people don't bother with moduli of convergence, because they just want to know that the limit exists or has a certain value, not how quickly it converges. But they come up in numerical analysis, where you want to know how much work it will take to approximate the limit to a desired tolerance; they come up in constructive analysis, where you use weaker logical principles and so can't always prove that a modulus exists (and so you're interested in the special cases where you can); and they come up in recursion theory, where you're interested in whether the modulus is computable.
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Very insightful. Thanks for the answer. I’m wondering if the converse is true: does every convergent sequence have a modulus of convergence? – mhdadk Jul 14 '23 at 00:40
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1Yes, that's the proof that I put in parentheses: since (given $\epsilon$) there's at least one $N$ that works in the definition of convergence, you can let $N(\epsilon)$ be the smallest such value of $N$. (Or you can use the axiom of choice to say that there must be some function picking one of these values of $N$ for each $\epsilon$.) Also notice that your statement ‘if $|a_n-c|\geq\epsilon$, then $n<N(\epsilon)$’ is equivalent to ‘if $n\geq N(\epsilon)$, then $|a_n-c|<\epsilon$’, so there's no difference from writing things in the other order. – Toby Bartels Jul 17 '23 at 07:59