As a sequence is the n^th root of a diagonal Ramsey number R(n, n) increasing, decreasing or neither?

Question

As a sequence is the n^th root R(n, n)^1/n of a diagonal Ramsey number R(n, n) increasing, decreasing or neither?

As a sequence of positive real numbers is the n^th root of a diagonal Ramsey number, bounded in [sqrt(2), 4], namely

R(n, n)^1/n

also decreasing as n --> oo if infinitely many terms of R(n, n)^1/n are in the compact set [sqrt(2), 4]?

Actually as an infinite sequence of real numbers is

R(n, n)^1/n

monotone at all (I mean either decreasing, increasing), or not monotone? If it is not monotone in some interval for infinitely many n it cannot converge. I mean there's tons of real analysis stuff on this.

A monotone increasing or monotone decreasing infinite sequence of real numbers with all or most of its terms inside a closed and bounded set converges (real analysis).

But I can find nothing on this in the graph theory literature regarding R(n, n)^1/n because the emphasis is on constructions and working with complete subgraphs K_n, and on finding finer upper bounds on the nth root of diagonal Ramsey numbers, using probabilistic methods like the Lovasz Local lemma, not on real number sequences.

I think I have shown correctly that, as n --> oo,

r(n + 1)^(1/n+1)/r(n)^1/n < 1,

but if you think this is in error or if you disagree please tell me why.

Answer 1

To the best of my knowledge, and see for example the respective Open Problems Graden page , it is still unknown whether $$\lim_{n \to \infty} R(n,n)^{1/n}$$ exists (but it seems plausible it does).

Any monotonicity result would imply the existence of this limit, as the sequence is known to be bounded (by classical results, as you mention).

Thus, also, the monotonicty question is (to the best of my knowledge) open.

I am unanware of any result directly contradicting your claim, so I do not disagree with it. However, it would be quite an important result, so my default stance is to be sceptical of any claims.

(ps: Please do not use this site for detailed inquiries on the correctness of your proof(s).)

Answer 2

It is not known that $R(n,n)^{1/n}$ is monotonic, it is not even known if $R(7,7)^{1/7}$ is smaller or bigger than $R(8,8)^{1/8}$.

If we knew that $R(7,7)^{1/7}$ is smaller than $R(8,8)^{1/8}$, then the known bound $R(7,7)\geq 205$ would imply $R(8,8)\geq 439$, whereas the best known lower bound is $R(8,8)\geq 282$.

If we knew that $R(7,7)^{1/7}$ is bigger than $R(8,8)^{1/8}$, then the known bound $R(7,7)\leq 540$ would imply $R(8,8)\leq 1326$, whereas the best known upper bound is $R(8,8)\leq 1870$.

Numbers taken from here.

Note that a sequence can be convergent without being monotonic or even resembling to be monotonic.

Finally, this site is not the right forum to advertise or check your work.