CalculusMediumInfinite Sequences and Series ShareInfinite Sequences and Series · Medium problemUse the ratio test on ∑n=1∞(2n)!(n!)2\displaystyle\sum_{n=1}^{\infty}\frac{(2n)!}{(n!)^{2}}n=1∑∞(n!)2(2n)!. State LLL and conclude.A(2n+2)(2n+1)(n+1)2→4>2\frac{(2n+2)(2n+1)}{(n+1)^2}\to4>2(n+1)2(2n+2)(2n+1)→4>2B(n+1)2(2n+2)(2n+1)→4>1\frac{(n+1)^2}{(2n+2)(2n+1)}\to4>1(2n+2)(2n+1)(n+1)2→4>1C(2n+2)(2n+1)(n+1)3→4>1\frac{(2n+2)(2n+1)}{(n+1)^{3}}\to4>1(n+1)3(2n+2)(2n+1)→4>1D(2n+2)(2n+1)(n+1)2→4>1\frac{(2n+2)(2n+1)}{(n+1)^2}\to4>1(n+1)2(2n+2)(2n+1)→4>1Check Show answer Walk through it step by step Practice more like this Review this concept