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.

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

#Combinatorics #NumberTheory #Mathematics

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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

Michael Mendrin - 7 years, 4 months ago

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

Vikram Pandya - 7 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Question about π2π^2π2 and Infinite Series

Comments

Question about $π^2$ and Infinite Series