\[\frac{{2\sqrt 2 }}{{9801}}\sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {4n} \right)!\left( {1103 + 26390n} \right)}}{{{{\left( {n!} \right)}^4}{{396}^{4n}}}}} = \frac{1}{\pi }.\]