[1sin(1/n)]/[n²+n+1]+[2sin(2/n)]/[n²+n+2]+……+[nsin

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/09 01:20:06

[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

证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1) 求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷高数求极限n趋于无穷大时,lim (1/n - sin(1/n))/ (1/n^2),lim (1/n - sin(1/ 2^n/n*(n+1) lim1/n(sin1/n+……+sin（n-1）/n)=?n趋向无穷大证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n 求极限：lim((2n∧2-3n+1)/n+1)×sin n趋于无穷紧急：求 lim n*sin(π(n^2+2)^0.5）*(-1)^n,n趋向无穷大； n→无穷大 sin^n(2nπ/3n+1)的极限怎么求解 n趋于正无穷求极限n^2*ln[n*sin(1/n)] (1/(n^2 n 1 ) 2/(n^2 n 2) 3/(n^2 n 3) ……n/(n^2 n n)) 当N越于无穷大级数收敛性之sin(1/n)>(2/π)×(1/n)