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# Question about $$π^2$$ and Infinite Series

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.

Note by Vikram Pandya
4 years, 1 month ago

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Hey that smart guy in Yahoo Answers may be onto something with the divisibility of (2n-2)! ./ (n-1)!(n-1)! by 3.

- 4 years, 1 month ago

Thanks for notifying ;-) I am glad that you notified about that smart guy

- 4 years, 1 month ago