The series \(\displaystyle 1-x^2+\frac{x^4}{2!} - \frac{x^6}{3!}+\frac{x^8}{4!} + ... + (-1)^n\frac{x^{2n}}{n!}+...\) converges to which of the following?

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The Maclaurin series for \(e^x\) is:

$$ 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!} + ... = \sum_{n=0}^\infty \frac{x^n}{n!}$$

Substituting \(-x^2\) results in the requested series:

$$ 1+\frac{-x^2}{1!}+\frac{(-x^2)^2}{2!}+\frac{(-x^2)^3}{3!} + ... = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!}$$ $$ = 1-x^2+\frac{x^4}{2!} - \frac{x^6}{3!}+\frac{x^8}{4!} + ... = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{n!}$$