# sum of power series

#### nngktr

This might seem easy, but could anyone share the trick how to do the following sums?
the sum of 8^n/(3n)!, n goes from 0 to infinity
the sum of 27^n/(3n+1)!, n goes from 0 to infinity
Any input is appreciated!

#### Opalg

MHF Hall of Honor
This might seem easy, but could anyone share the trick how to do the following sums?
the sum of 8^n/(3n)!, n goes from 0 to infinity
Let $$\displaystyle \omega$$ be a complex cube root of 1. Then $$\displaystyle \sum_{n=0}^\infty\frac{2^n}{n!} = e^2,\quad \sum_{n=0}^\infty\frac{(2\omega)^n}{n!} = e^{2\omega},\quad \sum_{n=0}^\infty\frac{(2\overline{\omega})^n}{n!} = e^{2\overline{\omega}}.$$

However, $$\displaystyle 1^n+\omega^n + \overline{\omega}^{\,n}$$ is equal to 3 if n is a multiple of 3, and 0 otherwise.

Therefore $$\displaystyle \sum_{n=0}^\infty\frac{8^n}{(3n)!} = \tfrac13\bigl(e^2 + e^{2\omega} + e^{2\overline{\omega}}\bigr).$$ The expression on the right should obviously be real, so you need to plug in the value for $$\displaystyle \omega$$ and check that this is indeed the case.