Morley's Formula

$$\sum_{k=0}^\infty {m\choose k}^3 = 1+\left({m\over 1}\right)^3+\left[{m(m+1)\over 1\cdot 2}\right]^3+\ldots$$

$$= {\Gamma(1-{\textstyle{3\over 2}} m)\over [\Gamma(1-{\textstyle{1\over 2}}m)]^3} \cos({\textstyle{1\over 2}}m\pi),$$

where ${n\choose k}$ is a Binomial Coefficient and $\Gamma(z)$ is the Gamma Function.