Question

这个数列n项和的极限怎么求。如图。

编辑不好，见谅。

Answer 1

L =lim(n->∞)∑(i:1->n) [ ( sin(iπ/n))/(n+1) ]

S = sin(π/n) + sin(2π/n)+...+ sin(nπ/n)
2cos(π/n) . S = 2sin(π/n).cos(π/n) + 2sin(2π/n).cos(π/n)+...+ 2sin(nπ/n).cos(π/n)
= [sin(2π/n)+sin0] +[sin(2π/n)+sin(π/n)]+...+[sin((n+1)π/n)+sin((n-1)π/n)]
= sin0 + sin((n+1)π/n)+ 2S -sin(π/n) - sin(nπ/n)
2(cos(π/n)+1)S = sin((n+1)π/n) -sin(π/n)
= 2cos[(n+2)π/(2n)]sin(π/2)
=2cos[(n+2)π/(2n)]
=2cos(π/2+π/n)
S =cos(π/2+π/n) / (cos(π/n)+1)

L =lim(n->∞)∑(i:1->n) [ ( sin(iπ/n))/(n+1) ]
= lim(n->∞) S/(n+1)
= lim(n->∞) cos(π/2+π/n) / [(n+1)(cos(π/n)+1)]
=0