May 30, 2021 - 15:04

An infinite series problem.

By Vamshi Jandhyala in mathematics

May 30, 2021

Problem

Solution

Let $1$, $\omega$ and $\omega^2$ be the cube roots of unity.

We have,

$$ \begin{aligned} e^1 &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots \\
e^{\omega} &= 1 + \omega + \frac{\omega^2}{2!} + \frac{\omega^3}{3!} + \dots \\
e^{\omega^2} &= 1 + \omega^2 + \frac{\omega^4}{2!} + \frac{\omega^6}{3!} + \dots \\
\end{aligned} $$

Adding the above three equations we get,

$$ \begin{aligned} \frac{1}{0!} + \frac{1}{3!} + \frac{1}{6!} + \dots &= \frac{e + e^\omega + e^{\omega^2}}{3} \\
&= \frac{e + e^{-\frac{1}{2} + i\frac{\sqrt{3}}{2}} + e^{-\frac{1}{2} - i\frac{\sqrt{3}}{2}}}{3} \\
&= \frac{e}{3} + \frac{2}{3\sqrt{e}} \cos \frac{\sqrt{3}}{2} \end{aligned} $$

Similarly, we have

$$ \begin{aligned} \frac{1}{2!} + \frac{1}{5!} + \frac{1}{8!} + \dots &= \frac{e + \omega e^\omega + \omega^2 e^{\omega^2}}{3} \\
\frac{1}{1!} + \frac{1}{4!} + \frac{1}{7!} + \dots &= \frac{e + \omega^2 e^\omega + \omega e^{\omega^2}}{3} \end{aligned} $$