Prove or disprove:

Let \[s(n) = \sum_{k=0}^∞ \frac{1}{{k+n \choose n}^{2}}\]

Then s(n) differs from an integral multiple of \(π^2\) by a rational number just in case the ternary representation of \(n-1\) has at least one \(2\) in it.

For instance for integer \(n, s(n)\) is always of the form \(a + bπ^2\), where a and b are rational numbers, but for \(n = 3,6 to 9, 12,15 to 27, 30..., b \) is an integer.

That is, for \( (n - 1) = 2, 12, 20, 21, 22, 102, 112, 120, ... \)in ternary (base 3) form b is an integer.

Source: http://answers.yahoo.com/question/index?qid=20140131085034AA68jaX

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TopNewestHey that smart guy in Yahoo Answers may be onto something with the divisibility of (2n-2)! ./ (n-1)!(n-1)! by 3. – Michael Mendrin · 3 years, 5 months ago

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– Vikram Pandya · 3 years, 5 months ago

Thanks for notifying ;-) I am glad that you notified about that smart guyLog in to reply