∑(n:1->m) sin(πn/2)k=1,2,....m =4k-3: ∑(n:1->m) sin(πn/2) =1m =4k-2: ∑(n:1->m) sin(πn/2) =1m =4k-1: ∑(n:1->m) sin(πn/2) =0m =4k: ∑(n:1->m) sin(πn/2) =1=> ∑(n:1->m) sin(πn/2) 发散
