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Factorial series

Evaluate the series

    \[\mathcal{S} = \sum_{n=0}^{\infty} \frac{1}{(3n)!}\]

Solution

Let \delta_{n, k} denote Kronecker’s delta and \zeta_m = e^{2\pi m i /3}. We have successively,

    \begin{align*} \sum_{n=0}^{\infty} \frac{1}{(3n)!} &=\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{\delta_{k, 3n}}{k!} \\ &=\frac{1}{2\pi i } \sum_{n=0}^{\infty} \sum_{k=0}^{\infty}\; \oint \limits_{\left | z \right |=R>1} z^{3n-k+1} \, \mathrm{d} z\\ &= \frac{1}{2\pi i }\oint \limits_{ \left | z \right |=R>1} \frac{1}{z} \sum_{n=0}^{\infty} \frac{1}{z^{3n}} \sum_{k=0}^{\infty} \frac{z^k}{k!} \, \mathrm{d} z\\ &= \frac{1}{2\pi i} \oint \limits_{\left | z \right |=R>1} \frac{e^z}{z \left ( 1-\frac{1}{z^3} \right )}\, \mathrm{d} z\\ &= \frac{1}{2\pi i } \oint \limits_{\left | z \right |=R>1} \frac{e^z z^2}{z^3-1} \, \mathrm{d}z \\ &=\sum_{m=-1}^{1}\frac{\zeta_m^2 e^{\zeta_m}}{3\zeta_m^2} \end{align*}

It follows that

    \[\mathcal{S} =\frac{e}{3} + \frac{2 \cos \frac{\sqrt{3}}{2}}{3\sqrt{e}}\]

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