sin(n*π/2)*sin(n*π/3)*sin(n*π/4)*...*sin(n*π/n-1) 求化简成一个关于n的表达式,

sin(n*π/2)*sin(n*π/3)*sin(n*π/4)*...*sin(n*π/n-1) 求化简成一个关于n的表达式, 令n趋近于无穷大,且n存在,求sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=?. lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=? 当n趋于无穷时,求[sin(π/n)/(n+1)+sin(2π/n)/(n+1/2)+.sinπ/(n+1/n)]的极限 sin(π/n)的收敛性 求Lim (2sin^n x+3cos^n x)∕(sin^n x+cos^n x) ,0≤x≤π/2n趋于无穷大 化简sin(nπ+2π/3)×cos(nπ+4π/3) n属于Z 判别级数∑(n=1,∝) 2^n sin(π/3^n) 的敛散性 sin(nπ+π/2n)条件收敛 级数(1/n) × sin(πn/2)的敛散性 级数(1/n) × sin(πn/2)的敛散性 求∑sin(n^2+1)π/n条件收敛 an=sin(nπ/2),n→无限,极限. 化简sin×[a+(2n+1)π]+2sin×[a-(2n+1)π]/sin(a-2nπ）coS(2nπ-a) (n属于Z) 化简sin(a+nπ)+sin(a+nπ)/sin(a+nπ)cos(a-nπ)(n∈z) {sin(θ+nπ)+sin(θ-nπ)}/sin(nπ+θ)cos(θ-nπ) n∈Z 化简 lim[sinπ/(√n＾2＋1)+sinπ/(√n^2+2)+．．．＋sinπ/√n^2+n),n—>无穷 证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1))