How do you use the ratio test to test the convergence of the series ∑ [n(n!)^2]/(2n+1)!n(n!)22n+1! from n=1 to infinity?

1 Answer
Jan 24, 2017

The series:

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

is convergent.

Explanation:

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.