Reciprocal of factorials

Let the sum be $S$ , and let $\omega=e^{2\pi i/3}$ , a complex cube-root of $1$ . Note that $\omega$ satisfies $\omega^2+\omega+1=0$ .

Consider the following sums: $\begin{aligned} S_0 &=\sum_{n=0}^{\infty} \frac{1}{n!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots \\ S_1 &=\sum_{n=0}^{\infty} \frac{\omega^n}{n!} =\frac{1}{0!}+\frac{\omega}{1!}+\frac{\omega^2}{2!}+\frac{1}{3!}+\frac{\omega}{4!}+\cdots \\ S_2 &=\sum_{n=0}^{\infty} \frac{\omega^{2n}}{n!}=\frac{1}{0!}+\frac{\omega^2}{1!}+\frac{\omega}{2!}+\frac{1}{3!}+\frac{\omega^2}{4!}+\cdots \end{aligned}$

When $n$ is a multiple of $3$ , $1+\omega^n+\omega^{2n}=3$ . When $n$ is not a multiple of $3$ , $1+\omega^n+\omega^{2n}=0$ . We cancel exactly the terms we wanted to! Adding the three sums, $S_0+S_1+S_2=3\sum_{n=0}^{\infty} \frac{1}{(3n)!}=3S$

But the sums are just exponentials in power-series form: $\begin{aligned} S_0&=e \\ S_1&=e^\omega=e^{-1/2+i\sqrt3/2}=e^{-1/2}\left( \cos \frac{\sqrt3}{2} + i \sin \frac{\sqrt3}{2} \right)\\ S_2&=e^{\omega^2}=e^{-1/2-i\sqrt3/2}=e^{-1/2}\left( \cos \frac{\sqrt3}{2} - i \sin \frac{\sqrt3}{2} \right) \end{aligned}$

Adding these, we have $S=\frac13 \left[e+\frac{2\cos \frac{\sqrt3}{2}}{\sqrt{e}} \right] = \boxed{1.168\ldots}$

Consider the Maclaurin series of $e^x$ , where $\omega$ denotes the cubic roots of unity. And note that $\omega^2 + \omega + 1 = 0$ and $\omega^3 =1$ .

$\begin{aligned} e^x & = \frac 1{0!} + \frac x{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + \frac {x^5}{5!} + \frac {x^6}{6!} + \frac {x^7}{7!} + \cdots \\ e^1 & = \frac 1{0!} + \frac 1{1!} + \frac 1{2!} + \frac 1{3!} + \frac 1{4!} + \frac 1{5!} + \frac 1{6!} + \frac 1{7!} + \cdots \\ e^\omega & = \frac 1{0!} + \frac \omega{1!} + \frac {\omega^2}{2!} + \frac 1{3!} + \frac \omega{4!} + \frac {\omega^2}{5!} + \frac 1{6!} + \frac \omega{7!} + \cdots \\ e^{\omega^2} & = \frac 1{0!} + \frac {\omega^2}{1!} + \frac \omega{2!} + \frac 1{3!} + \frac {\omega^2}{4!} + \frac \omega{5!} + \frac 1{6!} + \frac {\omega^2}{7!} + \cdots \end{aligned}$

$\begin{aligned} \implies e + e^\omega + e^{\omega^2} & = \frac 3{0!} + \frac 3{3!} + \frac 3{6!} + \frac 3{9!} + \cdots \\ \implies \sum_{n=0}^\infty \frac 1{(3n)!} & = \frac {e + e^\omega + e^{\omega^2}}3 & \small \blue{\text{Note that }\omega = e^{\frac {2\pi}3i}= e^{-\frac 12 +\frac {\sqrt 3}2i }} \\ & = \frac {e+e^{-\frac 12}(e^{\frac {\sqrt 3}2i}+e^{-\frac {\sqrt 3}2i})}3 & \small \blue{\text{and }\omega^2 = - \frac 12 - \frac {\sqrt 3}2i} \\ & = \frac {e+2e^{-\frac 12}\cos \frac {\sqrt 3}2}3 \\ & \approx \boxed{1.168} \end{aligned}$

Observe that: $e^{\frac{2\pi i}{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ $(e^{\frac{2\pi i}{3}})^2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$ $(e^{\frac{2\pi i}{3}})^3=1$ Now: $e^{-\frac{1}{2}+i\frac{\sqrt{3}}{2}}=\sum_{j=0}^{\infty}\frac{(-\frac{1}{2}+i\frac{\sqrt{3}}{2})^j}{j!}=\orange{1}+\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+\frac{-\frac{1}{2}-i\frac{\sqrt{3}}{2}}{2!}+\red{\frac{1}{3!}}+\frac{-\frac{1}{2}+i\frac{\sqrt{3}}{2}}{4!}+...$ $e^{-\frac{1}{2}-i\frac{\sqrt{3}}{2}}=\sum_{j=0}^{\infty}\frac{(-\frac{1}{2}-i\frac{\sqrt{3}}{2})^j}{j!}=\orange{1}+\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)+\frac{-\frac{1}{2}+i\frac{\sqrt{3}}{2}}{2!}+\red{\frac{1}{3!}}+\frac{-\frac{1}{2}-i\frac{\sqrt{3}}{2}}{4!}+...$ $e^{1}=\sum_{j=0}^{\infty}\frac{1}{j!}=\orange{1}+1+\frac{1}{2!}+\red{\frac{1}{3!}}+\frac{1}{4!}+...$

We can observe that the imaginary parts of the complex numbers will cancel out and the real parts add up to -1 which will cancel out with the terms of the taylor series of $e$ . This leaves us with the reciprocals of the factorials of the multiples of three added three times so:

$\sum_{n=0}^{\infty}\frac{1}{(3n)!}=\frac{e^{-\frac{1}{2}+i\frac{\sqrt{3}}{2}}+e^{-\frac{1}{2}-i\frac{\sqrt{3}}{2}}+e}{3}=\frac{e^2+2 \sqrt{e} \cos{\frac{\sqrt{3}}{2}} }{3e}\approx\boxed{1.168}$

In general:

$\sum_{n=0}^{\infty} \frac{1}{(mn)!}= \frac{1}{m} \sum_{k=1}^{m} e^{\omega^{k}}: \omega =e^{\frac{2\pi i}{m}}, m\in\mathbb{Z}>0$

Let $A(x)=\sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}$ .

Then it follows that $A'''(x)=A(x), A(0)=1, A'(0)=A''(0)=0$ .

This differential equation has general solution: $A(x)=C_1e^x+C_2e^{\zeta x}+C_3e^{\zeta^2 x}$ , where $\zeta$ is a non-real third root of unity.

Substituting the boundary conditions gives $C_1+C_2+C_3=1$ , $C_1+\zeta C_2+\zeta^2 C_3=0$ , and $C_1+\zeta^2 C_2+\zeta C_3=0$ . Solving these gives $C_1=C_2=C_3=\frac{1}{3}$ , and hence: $A(x)=\frac{1}{3}(e^x+e^{\zeta x}+e^{\zeta^2 x})$ .

Now substituting the value $x=1$ yields the solution to the problem: $A(1)=\frac{1}{3}(e+e^{\zeta}+e^{\zeta^2})\approx \boxed {1.17}$ .