Question

高数，常数项级数敛散性的判断

Answer 1

a = kπ (k 为整数)时， sina = 0，级数收敛。
a ≠ kπ (k 为整数)时，sina ≠ 0，
ρ = lim<n→∞>a<n+1>/a<n>
= lim<n→∞> (n+1)! n^n/[(n+1)^(n+1) n!]
= lim<n→∞> n^n/[(n+1)^n] = lim<n→∞> [n/(n+1)]^n
= lim<n→∞> [1-1/(n+1)]^n = lim<n→∞> {[1-1/(n+1)]^[-(n+1)]}^[-n/(n+1)]
= e^lim<n→∞>[-n/(n+1)] = 1/e < 1, 级数收敛，绝对收敛。