=1-cos[π/(2n)]
2sin[π/(4n)]·sin[3π/(4n)]
=cos[π/(2n)]-cos[2π/(2n)]
2sin[π/(4n)]·sin[3π/(4n)]
=cos[2π/(2n)]-cos[3π/(2n)]
……
2sin[π/(4n)]·sin[(2n-1)π/(4n)]
=cos[(n-1)π/(2n)]-cos[nπ/(2n)]
全部加起來，得到
2sin[π/(4n)]·∑sin[(2k-1)π/(4n)]
=1-cos[2π/(2n)]
=1
∴∑sin[(2k-1)π/(4n)]=1/{2sin[π/(4n)]}
令t=π/(4n)，則
原式=lim(t→0)π?/(16t?)·(1-t/sint)
=π?/16·lim(t→0)(sint-t)/(t?sint)
=π?/16·lim(t→0)(sint-t)/t?
=π?/16·lim(t→0)(cost-1)/(3t?)
=π?/16·(-1/6)
=-π?/96