Use the Ratio test to determine whether the series is Convergent or Divergent? Plz help me find S_n

1 + 2/(1*2*3) + 3/(1*2*3*4*5) + ...

I just need help finding the S_n part... I know that the numerator will be n but I don't understand how to get the denominator with all of these ! (whatever they're called).

Thanks a lot!

S_n = n / (2n - 1)!

By Ratio Test

Lim (n->INF) [(n+1) / (2(n+1) - 1)!] / [n / (2n - 1)!]

Lim (n->INF) [(n+1) / (2n + 1)!] / [n / (2n - 1)!]

Lim (n->INF) [(n+1) / (2n + 1)!] * [(2n - 1)! / n]

Lim (n->INF) [(n + 1)(2n - 1)!] / [n(2n + 1)!]

Lim (n->INF) [(n+1)(2n-1)!] / [n(2n+1)(2n)(2n - 1)!]

Lim (n->INF) [n + 1] / [n(2n + 1)(2n)]

Lim (n->INF) [n + 1] / [4n^3 + 2n^2]

Lim (n->INF) n(1 + 1/n) / n(4n^2 + 2n)

Lim (n->INF) (1 + 1/n) / (4n^2 + 2n) = [1 + 0] / INF = 0

Since this limit is less than 1 we know the series converges.

General term is n/(2n-1)!
Using the ratio test
r_n+1/r_n = (n+1)/(2n+1)! * (2n-1)!/n
= (n+1)/(n(2n)(2n+1)) = (n+1)/(2n^2(2n+1))
Limit as n->infinity is 0