lim n→∞(sin(π/n))∑(1/(1+cos(k/n))) = 其中k=1~n

lim（ n→∞）(sin(π/n))∑(1/(1+cos(k/n)))
= lim （n→∞）(π/n)∑(1/(1+cos(k/n)))
=π∫[0,1]1/(1+cosx)dx
=π∫[0,1](1-cosx)/(1-cos²x)dx
=π∫[0,1](1/sin²x-cosx/sin²x)dx
=π(-cotx+1/sinx)|[0,1]
=π(1-cosx)/sinx|[0,1]
=π[(1-cos1)/sin1-lim(x->0)(1-cosx)/sinx]
=π[(1-cos1)/sin1-1/2]