Question

高等数学，求大佬

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Answer 1

(1) ∑<n=1,∞> sin(π/4^n) < ∑<n=1,∞> π/4^n ,
后者为公比是 1/4 的等比级数收敛，则原级数收敛。
(2) ∑<n=1,∞> 1/[n√(n+1)]< ∑<n=1,∞> 1/n^(3/2),
后者为p-收敛，p = 3/2 > 1, 收敛, 则原级数收敛。
(3) ∑<n=1,∞> (n+2)/3^n
ρ = lim<n→∞>a<n+1>/a<n> = lim<n→∞>(n+2)3^n/[(n+1)3^(n+1)]
= lim<n→∞>(n+2)/[3(n+1)] = lim<n→∞>(1+2/n)/[3(1+1/n)] = 1/3 < 1,
级数收敛。
(4) ∑<n=1,∞> n!/(2^n+1)
ρ = lim<n→∞>a<n+1>/a<n> = lim<n→∞>(n+1)!(2^n+1)/{n![2^(n+1)+1]}
= lim<n→∞>(n+1)(1+1/2^n)/[2+1/2^n] = +∞, 级数发散。

Answer 2

大学毕业好几年，高数知识已经还给高数老师了。

Answer 3