What is the sum of this series: $\displaystyle\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$ I tried getting it into a closed form but failed. Could someone help me out?
$$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$

You should know the Taylor series for $e^x$: 
$e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \cdots$
Then, the Taylor series for $e^{-x}$ is: 
$e^{-x} = 1 + (-x) + \dfrac{(-x)^2}{2} + \dfrac{(-x)^3}{3!} + \dfrac{(-x)^4}{4!} + \cdots$
$e^{-x} = 1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{3!} + \dfrac{x^4}{4!} - \cdots$
Add these two series together, and all the odd terms cancel. So you get: 
$e^x+e^{-x} = 2 + 2\dfrac{x^2}{2} + 2\dfrac{x^4}{4!} + \cdots$
$\dfrac{e^x+e^{-x}}{2} = 1 + \dfrac{x^2}{2} + \dfrac{x^4}{4!} + \cdots$
Now, plug in $x = \sqrt{a}$, and see what you get. 
