Question

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?

Answer 1

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!