Question

can someone please show me how to use the ratio test for sigma n=1 to infinity (n!)^2/(2n!)

Answer 1

taking a ratio of two terms:
a(n+1)/a(n) by expanding factorial terms:
(n+1)^2 / [ (2n+2)(2n+1) ]
= (n^2 + 2n + 1) / ( 4n^2 + 6n + 2)

Letting n -> inf,
a(n+1)/a(n) approaches 1/4 < 1 and therefore converges.