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1.16


\begin{proof}We have \begin{displaymath}\left({{n}\over{k(n-k)}}\right)^{1/2} ... ... \cdot e^{k^2/n}\sqrt{k} = \sqrt{2\pi k}\cdot e^{k^2/n + 1/6k}.$ \end{proof}

Jukna 2003-01-15