{[(-1)^(n+1)]/n}*sin(nx),n=1,2,3….求n从0到正无穷求和 matlab的code求的是对于对于整体x从-pi到pi的积分。我刚才写错了

|sin(1/n)| 求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷 lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=? 求极限（sin^2)n/n+1 证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1)) lim[nsin(1/n)+1/n(sin n)]= lim（1/n）sin n （n→∞） 当n趋向于无穷大时,n{sin(1/n)-sin[1/(n+1)]}趋向 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, lim1/n(sin1/n+……+sin（n-1）/n)=?n趋向无穷大 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 极限 lim(x-->无穷)(1/n sin(n)+1/n sin(1/n)+nsin(1/n))求极限lim(x-->无穷)(1/n sin(n)+1/n sin(1/n)+nsin(1/n)) [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? 求sin(√(n+1))-sin(√n),当n趋向无穷的极限 当n趋于∞,[(n+1)^(n+1)/n^n]sin(1/n)的极限怎么求 (sin(ln2/2)+sin(ln3/3)+...+sin(lnn/n))^（1/n）的极限 ((-1)^(n-1))/(n+1)*sin(n!),当n趋向无穷大时的极限 ∏(k从1到n-1)sin(kπ/n) = n / 2^(n-1)