Proof that asymptotic density $>1/n$ implies every sufficiently large integer is the sum of $n$ terms

Question

Gerry Myerson commented on a previous question, which at the time asked for proof that every integer is a sum of two deficient numbers, "The deficient numbers have natural density strictly greater than one-half. Isn't that enough to prove that every sufficiently large integer is a sum of two deficients?"

This seems like an intuitive claim, but I'm having trouble finding a proof of it, or of the more general claim that density $>1/n$ implies every sufficiently large integer is the sum of $n$ terms of the set.

Edit:

I was not consistent in my previous statement of the question. Previously the title read "a sum of at most $n$ terms", but the body lacked the "at most" condition. The title has been changed to reflect the body. Further, I have altered the question to require the density to be strictly greater than $1/n$. I realize also that asking for a more general result than the one I had significant reason to believe has been proven is somewhat of an abuse of the reference request tag, but given that references have been provided which are thought to contain the result I will leave it in this instance. I do not fully understand the subtleties of the question as it might have been interpreted previously, so forgive me if I have invalidated any previously made comments or answers.

Lemma: Let $n$ be an integer greater than $1$, and let $S$ be a set containing more than half of all positive integers less than $n$. Then there exist $a,b\in S$ such that $a+b=n$.

Proof: Suppose first that $n$ is odd. Then every integer $a\geq\dfrac{n+1}2$ has a partner $b\leq\dfrac{n-1}2$ distinct from $a$ such that $a+b=n$, thus any selection of more than $\dfrac{n-1}2$ positive integers less than $n$ contains a pair which sums to $n$. Now suppose $n$ is even. Then every integer $a\geq\dfrac{n}2+1$ has a partner $b\leq\dfrac{n}2-1$ distinct from $a$ such that $a+b$, and we have one additional pair in which both terms are $\dfrac{n}2$, thus any selection of more than $\dfrac{n}2-1$ positive integers less than $n$ contains a pair (not necessarily distinct) which sums to $n$. Both $\dfrac{n-1}2+1$ and $\dfrac{n}2$ are greater than $\dfrac{n-1}2$, so the proof is finished.

Answer 1

As you showed in the edit to your question, if a set $S$ of natural numbers has density $\gt1/2$ (or more generally, has lower density $\gt1/2$, the density need not exist), then every sufficiently large natural number is the sum of two (distinct) elements of $S$.

The more general claim is false for all $n\gt2$, because the set $S$ of all even numbers has density $\gt1/3$.

Answer 2

It seems Mann's result does not suffice, it is amazingly sensitive to initial values. You want Kneser's result, short English discussion in

Nathanson, Melvyn B. (1990). "Best possible results on the density of sumsets". In Berndt, Bruce C.; Diamond, Harold G.; Halberstam, Heini et al. Analytic number theory. Proceedings of a conference in honor of Paul T. Bateman, held on April 25-27, 1989, at the University of Illinois, Urbana, IL (USA). Progress in Mathematics 85. Boston: Birkhäuser. pp. 395–403.

I will see what else is available....Kneser may not be good enough either. Wait on Gerry.

EEDDIITT: you might check Terence Tao and Van Vu, Additive Combinatorics, Cambridge University Press 2006. This should have your result and much more, although perhaps not in elementary language.