Solving Series Expansion Equations for pi^2/3! to pi^6/7!

[kevi555]

Hi there,

I'm looking for a series that expands to look like : (pi^2)/3! - (pi^4)/5! + (pi^6)/7! - ... =1

Any ideas would be greatly appreciated as I can't seem to get anywhere with it!

Thanks,


K

 

[CompuChip]

Can you start by writing the expression as a sum, i.e.
$$\pi^2 / 3! - \pi^4 / 5! + \pi^6 / 7! - \cdots = \sum_{n = n_0}^{N} a_n$$

Also, the Taylor series of the sine function
$$\sum_{n = 0}^\infty (-1)^n \frac{x^{2n+1}}{(2n + 1)!} = \sin(x)$$
may come in handy.