deletions | additions       diff --git a/Random.tex b/Random.tex  index 7d76c8e..0187e25 100644  --- a/Random.tex  +++ b/Random.tex 
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\end{equation}  Is there something special about $12$ then, such that \begin{equation}  \prod_{n=0}^{\infty}\frac{e^{12n}}{n!} = 1  \end{equation}  But then, it must be true, as $log(1)=0$ that \begin{equation}  \sum_{n=1}^\infty 12n - log(n!) =0  \end{equation}  Which can be rewritten as \begin{equation}  12\sum_{n=1}^\infty n = \sum_{n=1}^\infty log(n!)  \end{equation}  However we know the sum on the left hand side to be $-1/12$ such that \begin{equation}  \sum_{n=1}^\infty log(n!) = -1  \end{equation} then...