[1sin(1/n)]/[n²+n+1]+[2sin(2/n)]/[n²+n+2]+……+[nsin(n/n)]/[n²+n+n]求n趋于无穷大时极限?

[1sin(1/n)]/[n²+n+1]+[2sin(2/n)]/[n²+n+2]+……+[nsin(n/n)]/[n²+n+n]求n趋于无穷大时极限?
[1sin(1/n)]/[n²+n+1]+[2sin(2/n)]/[n²+n+2]+……+[nsin(n/n)]/[n²+n+n]
求n趋于无穷大时极限?

[1sin(1/n)]/[n²+n+1]+[2sin(2/n)]/[n²+n+2]+……+[nsin(n/n)]/[n²+n+n]求n趋于无穷大时极限?
由定积分基本定理可知原式 ＝求和符号（i=1,n趋值于无穷大）[n*(i/n)*sin(i/n)]/[n^2+n+n*(i/n)] =n*{积分符号(0到1)[(x*sinx)/(n+1+x)]dx} 因为n趋值于无穷大所以分数线的下面就可以直接写成n,因此 =积分符号(0到1)(x*sinx)dx 采用分步积分法 =(-xcosx+sinx){x=1,x=0} =sin1-cos1