Does $\sum_{n=1}^\infty \frac{T(n)}{2^{2^n}}$ converge?

Question

Let $T(n)$ be the number of distinct topologies on a set with $n$ elements. Does $\displaystyle\sum_{n=1}^\infty \displaystyle\frac{T(n)}{2^{2^n}}$ converge?

There is not much context to this unfortunately. It's a problem I came up with myself, when counting the number of topologies on an $n$-element set for $n=2,3$ (I am a beginner in topology). I am not sure of the difficulty of this problem but any progress toward a solution would be appreciated.

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Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you [edit] your question to include some motivation, and an explanation of your own attempts. — GNUSupporter 8964民主女神地下教會, May 13 '18 at 20:18
I didn't, but I suspected that would be the way forward, though it is very hard to estimate such a quantity. — EllipticCurve, May 13 '18 at 20:42
@GNUSupporter: I agree with your comment as a rule, but this is a question of obvious interest that is obviously not homework. — Rob Arthan, May 13 '18 at 21:03
@RobArthan When I commented, it's a PSQ. Whether this is HW or not is irrelevant. — GNUSupporter 8964民主女神地下教會, May 13 '18 at 21:08
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@GNUSupporter: What does PSQ mean? Whether or not the question is homework makes a very significant difference to my willingness to help. — Rob Arthan, May 13 '18 at 21:16
@RobArthan You may refer to this meta question for the meaning of PSQ, as well as How to ask HW questions on [meta]. It's OK to ask HW questions here. — GNUSupporter 8964民主女神地下教會, May 13 '18 at 21:26
@GNUSupporter: Thank you for the pointer to explain your obscure MSE TLA. I know fine well that it is OK to ask homework questions on MSE, but my opinion on whether a question is homework makes a significant difference to my willingness to help if the OP has given limited evidence of their own work. — Rob Arthan, May 13 '18 at 21:36

score 6 · Accepted Answer · answered May 13 '18 at 20:24

6

I believe the answer is yes. From this paper: The Number of Finite Topologies- D. Kleitman and B. Rothschild, the authors show that $T(n)$ is like $O(2^{n^2/4})$.

answered May 13 '18 at 20:24

Alex R.

33,289

With this fact that it's just the same as the number of preorders I suppose it doesn't require the full on analysis, as even with just binary relations there's only $2^{n^2}$. – Countingstuff May 13 '18 at 21:44

Does $\sum_{n=1}^\infty \frac{T(n)}{2^{2^n}}$ converge?

1 Answers1