need help with a series?

Question

does the series

summation n=1 to infinity of 1/((1)(3)(5)......(2n-1)) converge or diverge and what test did you use to determine it.

Answer 1

Since Sum (i=1 to ∞ of) 1/n! converges to e-1,

Your sum
= Sum (i=1 to ∞ of) (n! * 2^n)/(2n!)
= Sum (i=1 to ∞ of) n!/((2n!)/2^n)
< Sum (i=1 to ∞ of) n!/(n!^2)
= Sum (i=1 to ∞ of) 1/n!
= e-1
so your original sum converges

Answer 2

The series is Sum (1 to infinity) a_n, where a_n is the inverse of the product of the first n odd numbers.

For every n >2 , 1 . 3 . 5.... . (2n -1) > 2 . 4 .....(2n -2) = 2^(n-1) . (n-1)! > (n-1)!. Therefore,

0 < a_n < 1/(n-1)| and since Sum 1/(n -1)! = e -1, it follows Sum a_n converges