Question

`sum_(n=1)^oo(n!)^2/((2n)!)` converges ?

Answer 1

Using Stirling's asymptotic approximation formula

`n! approx sqrt(2pin)(n/e)^n` we have

`a_n = (n!)^2/((2n)!) approx sqrt(pi n) 1/2^(2n) < pi n 2^(-2n)` which converges

so

`sum_(n=1)^oo a_k` converges.

NOTE

`n x^(-2n)= -1/2 x d/(dx)x^(-2n)`

and `sum_(k=0)^oo x^(-2k) = x^2/(x^2-1)` for `abs x > 1` then

`sum_(n=0)^oo n x^(-2n) = -1/2xd/(dx) 1/(1-x^2) =( x(1+x^2))/(x^2-1)^2`

and then

`sum_(k=1)^oo a_n le pi (2(1+2^2))/(2^2-1)`