converge or diverge!!

With $a_{n} = \dfrac{1}{(2n + 1)!}$ we have by the ratio test that

$\lim_{n \rightarrow \infty} |\dfrac{a_{n+1}}{a_{n}}| = \lim_{n \rightarrow \infty} \dfrac{\frac{1}{(2n + 3)!}}{\frac{1}{(2n + 1)!}} = \lim_{n \rightarrow \infty} \dfrac{1}{(2n + 3)(2n + 2)} = 0,$

and thus the series is absolutely convergent, (and hence convergent).

We can also evaluate this series exactly. Note that the Maclaurin series for $\sinh(x)$ is

$\displaystyle\sum_{n=0}^{\infty} \dfrac{x^{2n + 1}}{(2n + 1)!}.$ Thus with $x = 1$ the series in this question has the sum $\sinh(1) - 1.$

The series $\sum_{n=1}^\infty \frac{1}{n!}$ is known to converge to $e - 1.$ The series in the problem contains strictly smaller terms greater than $0.$ By the comparison test, the series converges (more specifically to a value between $0$ and $e-1$ ).