The given sequences are;(2n)! 4" n=1Σ(-1)n+1, n=1 n(n+1)We will check the convergence and divergence of each sequence and justify the answers accordingly.(i) Convergence of (2n)! 4" n=1For the sequence given by (2n)! 4" n=1 to converge, the ratio test may be used. In order to use the ratio test, we must first find the sequence's terms: a_n=(2n)! 4" The ratio test yields;lim|a_n+1/a_n|= lim|[(2(n+1))! 4"]/[(2n)! 4"]|= lim|(2(n+1))(2n+1)16|= lim|(4n²+6n+2)|/16= ∞Since the limit is greater than 1, the sequence diverges.(ii) Convergence of Σ(-1)n+1, n=1 n(n+1)The given sequence can be proven to converge using the alternating series test. In order to do so, the sequence must first satisfy the two conditions of the test. For the series Σ(-1)n+1, n=1 n(n+1),a_n=n(n+1), which is a decreasing function for all n. Additionally, a_n approaches 0 as n approaches infinity.Thus, the sequence is convergent by the alternating series test.