Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.