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Assume $G = p/q$ with $q > 1$ where $p$ and $q$ are coprime. Let $k$ be an arbitrary large odd integer s.t. $2k + 1 > q$ and $p/q$ is not a rational multiple of $$s_k = \sum_{n = 0}^k \frac{(-1)^n}{(2n + 1)^2} = 1 - \frac{1}{3^2} + \frac{1}{5^2} - \ldots - \frac{1}{(2k + 1)^2}.$$ For odd $k$, $p/q - s_k$ is negative and any positive real number must greater than $p/q - s_k$. Note that $1/(2k + 1)!^2 > 0$, so we have $$-\ell < G - s_k < \frac{1}{(2k + 1)!^2}$$ or $$-(2k + 1)!^2\ell < (2k + 1)!^2\left( \frac{p}{q} - s_k \right) < 1$$ where $\ell$ is some integer. Now $(2k + 1)!^2(p/q - s_k)$ is not an integer because $p/q$ is not a rational multiple of $s_k$ by our choice of $k$. However, note that $q, 3^2, 5^2, \ldots, (2k + 1)^2$ must all be divisors of $(2k + 1)!^2$, i.e., \begin{align} (2k + 1)!^2\left( \frac{p}{q} - s_k \right) &= \frac{(2k + 1)!^2 p}{q} - (2k + 1)!^2 + \frac{(2k + 1)!^2}{3^2}\\ &\quad -\frac{(2k + 1)!^2}{5^2} + \cdots + \frac{(2k + 1)!^2}{(2k + 1)^2} \end{align} is an integer, a contradiction.

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Any rational number is a rational multiple of any other (non-zero) rational number. – Arthur Sep 2 '12 at 19:58
You're right. Is there any way to modify this proof? – glebovg Sep 2 '12 at 20:03
How did the factorials get into your expressions? We can get $|G-s_k|\lt \frac{1}{(2k+3)^2}$, but nothing substantially better. As to modifying the proof, that is quite a challenge, since whether the Catalan constant is irrational is a long-standing open problem. – André Nicolas Sep 2 '12 at 21:02
In fact, Mick has detected the main problem with this (incorrect) proof. I am trying a proof of irrationality for Catalan's constant via hypergeometric 3F2 functions, as you can check at arXiv ( ), but a definitive proof remains out of reach. Any comment on this line of attack is welcome. F. M. S. Lima (fabio -at- – Fabio M. S. Lima Sep 15 '12 at 21:22
Moreover, for odd integer $k$, $k>0$, $G-s_k$ is positive, contrarily to what you are assuming! – Fabio M. S. Lima Sep 15 '12 at 22:13
up vote 1 down vote accepted

You write assume the fraction p/q. Also p/q is not a rational multiple of a partial sum. But that partial sum is a rational number. Hence p/q is a rational multiple since any rational is a rational multiple of any other.

Contradiction and thats whats wrong.

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As Arthur wrote in a comment two hours earlier, right? – Gerry Myerson Sep 2 '12 at 23:49
I Think i wrote it more clearly ? – mick Sep 8 '12 at 9:19

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