求如下极限

lim(n→∞)∑√(n^2-i^2),i∈[0,n]
=lim(n→∞)n*∑√[1-(i/n)^2],i∈[1,n]
=lim(n→∞)n*∫√(1-(i/n)^2)di,i∈[0,n]

令 x=i/n,则 i=n*x,x∈[0,1]
=lim(n→∞)n*∫√(1-(x)^2)dnx,x∈[0,1]
=lim(n→∞)n^2*∫√(1-(x)^2)dx,x∈[0,1]

令x=sint,t∈[0,π/2]则dx=costdt,则
原式＝lim(n→∞)n^2*∫(cos t)^2)dt,t∈[0,π/2]
＝lim(n→∞)n^2*∫1/2*(1-cos 2t)dt,t∈[0,π/2]
＝lim(n→∞)n^2*1/2*π/2
＝lim(n→∞)n^2×π/4
显然该极限发散。分析原式为无穷个大于1的正数求和，显然发散，可能楼主抄题的时候漏除了n^2，即

lim(n→∞) 原式/n^2＝π/4

为无穷大
因
Lim_n->oo[sigma(i,form 0 ton)(n^2-i^2)^0.5]
＞ Lim_n->oo〔(n^2-0^2)^0.5]
＝Lim_n->oo〔n〕

lim(n→∞)∑√(n^2-i^2),i∈[1,n]
=lim(n→∞)∑√[1-(i/n)^2]/n,i∈[1,n]
=∫√(1-x^2)dx,x∈[0,1]
令x=sint,t∈[0,π/2]则dx=costdt，即
lim(n→∞)∑√(n^2-i^2),i∈[1,n]
=∫(cost)^2dt,t∈[0,π/2]
=(1/2)∫(1+cos2t)dt,t∈[0,π/2]
=[t+(sin2t)/2]/2,t∈[0,π/2]
=π/4
即lim(n→∞)∑√(n^2-i^2)=π/4,i∈[1,n]

附：该极限是切分单位圆在第一象限的面积得到的。

657647737