How Do You Apply The Ratio Test To Determine If Sigma (n!)/(1*3*5 ...

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? Calculus Tests of Convergence / Divergence Ratio Test for Convergence of an Infinite Series

1 Answer

Noah G 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!

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