2sin[π/(4n)]·sin[π/(4n)]
=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)]
全部加起来，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)，则
原式=lim(t→0)π²/(16t²)·(1-t/sint)
=π²/16·lim(t→0)(sint-t)/(t²sint)
=π²/16·lim(t→0)(sint-t)/t³
=π²/16·lim(t→0)(cost-1)/(3t²)
=π²/16·(-1/6)
=-π²/96
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