Catalan constant is irrational. What is wrong with this proof? 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.

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.
