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!
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.