The [tex]n[/tex]-th term [tex](n\ge1)[/tex] in the series is evidently
[tex]\dfrac{2n-1}{(2n-1)!} = \dfrac{2n-1}{(2n-1)(2n-2)!} = \dfrac1{(2n-2)!}[/tex]
By the ratio test, the infinite series converges, since
[tex]\displaystyle \lim_{n\to\infty} \left| \frac{\frac1{(2(n+1)-2)!}}{\frac1{(2n-2)!}}\right| = \lim_{n\to\infty} \frac{(2n-2)!}{(2n)!} = \lim_{n\to\infty} \frac1{2n(2n-1)} = 0 < 1[/tex]
