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
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
#L = (2 + 0 - 0)/(0 + 0)#
#L = 2/0#
#L = oo#
Since
Hopefully this helps!
