How do you apply the ratio test to determine if Sigma (n!)/(1*3*5* * *(2n-1)) from n=[1,oo) is convergent to divergent?

1 Answer
Jun 18, 2017

The series diverges.

Explanation:

This sum is represented by the formula

a_n = (n!)/(2n - 1)

The ratio test would be helpful here, because we're dealing with a fraction that involves factorials.

L = lim_(n->oo) (((n + 1)!)/(2(n + 1) - 1))/((n!)/(2n - 1))

Now we simplify.

L = lim_(n->oo) (((n + 1)!)/(2n + 1))/((n!)/(2n - 1))

L = lim_(n->oo) ((n + 1)!)/(2n + 1) * (2n - 1)/(n!)

L = lim_(n->oo) ((n + 1)(n!))/(2n + 1) * (2n - 1)/(n!)

L = lim_(n->oo) ((n + 1)(2n - 1))/(2n + 1)

L = lim_(n->oo) (2n^2 +n - 1)/(2n + 1)

Divide by the highest power.

L = lim_(n->oo) ((2n^2 + n - 1)/n^2)/((2n + 1)/n^2)

L = lim_(n->oo) (2 + 1/n - 1/n^2)/(2/n + 1/n^2)

We know that lim_(n->oo) 1/x = lim_(n->oo) 1/x^2 = 0. Hence,

L = (2 + 0 - 0)/(0 + 0)

L = 2/0

L = oo

Since L <1, we know that this sequence diverges and therefore doesn't have a finite sum.

Hopefully this helps!