How do you use the ratio test to test the convergence of the series #∑ [n(n!)^2]/(2n+1)!# from n=1 to infinity?

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Answer 1 · 2017-01-24T21:05:41

The series:

#sum_(n=1)^oo (n(n!)^2)/((2n+1)!)#

is convergent.

We have to analyse:

#lim_(n->oo) abs (a_(n+1)/a_n)#

so we first calculate the ratio:

#abs (a_(n+1)/a_n) = frac ( ( ( n+1) ((n+1)!)^2)/((2(n+1)+1)!)) ( (n(n!)^2)/((2n+1)!)) = (n+1)/n ((n+1)!)^2/(n!)^2 ((2n+1)!)/((2n+3)!)#

#abs (a_(n+1)/a_n) = (n+1)^3/(n(2n+3)(2n+2)) = (n^3+3n^2+3n+1)/(4n^3+10n^2+6n)#

and we can see that:

#lim_(n->oo) abs (a_(n+1)/a_n) =1/4 < 1#

which means the series is convergent.