How to find the Maclaurin series

Question

How can I find the Maclaurin series.

$\sum _{ k=0 }^{ \infty }{ ((-1)^ k }{\pi}^{2k}$/($9^k(2k)!$)

Please, help me how to solve this. Show how to do this.

Answer 1

Hint:

Remember that $$\cos u=\sum_{k\ge 0}(-1)^k\frac{u^{2k}}{(2k)!}.$$

Some details

In the present case, we have to identify $\dfrac{\pi^2}{9k}$ with $u^{2k}$, which yields $u^{2k}=\\Bigl(\dfrac\pi3\Bigr)^{2k}$, so the sum of the series is $\cos\dfrac\pi3$.