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.