对角Ramsey数$r(n, n)$的$n^{th}$根收敛性的非构造性证明
A nonconstructive Proof to show the Convergence of the $n^{th}$ root of diagonal Ramsey Number $r(n, n)$
摘要：
对角Ramsey数的$n^{th}$根收敛于有限极限吗？答案是肯定的。如果一个序列满足单调性以外的收敛条件，则它可以被证明是收敛的。我们证明了这样一个属性，其中序列的$n^{th}$根不收敛，即使没有先验知识，序列是否单调或不。我们还证明了对角Ramsey数的$n^{th}$根可以表示为两个因子的乘积，第一个因子是一个已知的收敛序列，第二个因子是一个绝对收敛的无穷级数。也可以表示为其中一个乘积是收敛的，而另一个乘积的所有值来自复平面上单位圆内的一致收敛复函数全纯。我们的动机仅仅是证明猜想作为一个问题，在寻找一个解决方案，而不是建立一些深刻的理论图。第二个问题是：如果极限存在，那么极限是什么？在写这篇文章的时候，我们的理解是，所寻求的证据不需要是建设性的。在这里，我们证明了非建设性的证明，$n^{th}$根的对角拉姆齐数收敛到一个有限的限制。我们还证明了对角Ramsey数的$j^{th}$根的极限是2，其中正整数$j$依赖于Ramsey数。
英文摘要：
Does the $n^{th}$ root of the diagonal Ramsey number converge to a finite limit? The answer is yes. A sequence can be shown to converge if it satifies convergence conditions other than or besides monotonicity. We show such a property holds for which the sequence of $n^{th}$ roots does converge, even if one has no a priori knowledge as to whether the sequence is monotone or not. We show also the $n^{th}$ root of the diagonal Ramsey number can be expressed as a product of two factors, the first being a known convergent sequence and the second being an absolutely convergent infinite series. One also can express it where one product is convergent and the other has all its values from a uniformly convergent complex function holomorphic within the unit disc on the complex plane. Our motivation solely is to prove the conjecture as a problem in search of a solution, not to establish some deep theory about graphs. A second question is: If the limit exists what is it? At the time of this writing the understanding is the proofs sought need not be constructive. Here we show by nonconstructive proofs that the $n^{th}$ root of the diagonal Ramsey number converges to a finite limit. We also show that the limit of the $j^{th}$ root of the diagonal Ramsey number is two, where positive integer $j$ depends upon the Ramsey number.
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