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.