We want to use the Ratio test, so we look at the limit:
lim n → ∞ | a n + 1 a n | = lim n → ∞ ( 2 n + 2 ) ! ( ( n + 1 ) 2 + 1 ) ( ( n + 1 ) ! ) 2 ( 2 n ) ! ( n 2 + 1 ) ( n ! ) 2 = lim n → ∞ ( 2 n + 2 ) ! ( n 2 + 1 ) ( n ! ) 2 ( 2 n ) ! ( ( n + 1 ) 2 + 1 ) ( ( n + 1 ) ! ) 2 = lim n → ∞ ( 2 n + 2 ) ! ( 2 n ) ! ⋅ ( n 2 + 1 ) ( n + 1 ) 2 + 1 ⋅ ( n ! ( n + 1 ) ! ) 2 = lim n → ∞ ( 2 n + 2 ) ( 2 n + 1 ) ⋅ n 2 + 1 n 2 + 2 n + 2 ⋅ 1 ( n + 1 ) 2 = lim n → ∞ 2 n + 2 n + 1 ⋅ 2 n + 1 n + 1 ⋅ n 2 + 1 n 2 + 2 n + 2 = 2 ⋅ 2 ⋅ 1 = 4 {\displaystyle {\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }{\frac {\frac {(2n+2)!}{((n+1)^{2}+1)((n+1)!)^{2}}}{\frac {(2n)!}{(n^{2}+1)(n!)^{2}}}}\\&=\lim _{n\to \infty }{\frac {(2n+2)!(n^{2}+1)(n!)^{2}}{(2n)!((n+1)^{2}+1)((n+1)!)^{2}}}\\&=\lim _{n\to \infty }{\frac {(2n+2)!}{(2n)!}}\cdot {\frac {(n^{2}+1)}{(n+1)^{2}+1}}\cdot \left({\frac {n!}{(n+1)!}}\right)^{2}\\&=\lim _{n\to \infty }(2n+2)(2n+1)\cdot {\frac {n^{2}+1}{n^{2}+2n+2}}\cdot {\frac {1}{(n+1)^{2}}}\\&=\lim _{n\to \infty }{\frac {2n+2}{n+1}}\cdot {\frac {2n+1}{n+1}}\cdot {\frac {n^{2}+1}{n^{2}+2n+2}}=2\cdot 2\cdot 1=4\end{aligned}}}
Since this limit is greater than 1 {\displaystyle 1} , by the Ratio test, the series ∑ n = 1 ∞ a n {\displaystyle \sum _{n=1}^{\infty }a_{n}} diverges.
, by the Ratio test, the series 
diverges.
