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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply