## Thomas' Calculus 13th Edition

Let us consider $a_n=\dfrac{(n!)^2}{(2n)!}$ Apply ratio test. $\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{((n+1)!)^2}{(2(n+1))!}}{\dfrac{(n!)^2}{(2n)!}}|$ $\implies \lim\limits_{n \to \infty}|\dfrac{n^2+1+2n}{4n^2+6n+2}|=\lim\limits_{n \to \infty}|\dfrac{1+\dfrac{1}{n^2}+\dfrac{2}{n}}{4+\dfrac{6}{n}+\dfrac{2}{n^2}}|=\dfrac{1+0+0}{4+0+0} =\dfrac{1}{4} \lt 1$ Thus, the series converges by the ratio test.