计算：underset(lim)(n→∞)(1)/(n)[sin(π)/(n)+sin(2π)/(n)+…+sin(（n-1）π)/(n)]．

【例1】极限lim_(ntoinfty)(2n+3sin n)/(n)=（）.A. ∞B. 2C. 3D. 5

__-|||-lim _(narrow infty )([ sin (dfrac {pi )(4)+dfrac (1)(n))] }^n=( )A.

_(n)=((-1))^n+1dfrac (1)(sqrt {n)}-|||-C. _(n)=sin dfrac (npi )(2)-|||-D. _(n)=d

(2) lim _(xarrow 0)dfrac (sin ({x)^n)}({(sin x))^m} (n,m为正整数),

(2) lim _(xarrow 0)dfrac (sin ({x)^n)}({(sin x))^m} (n,m为正整数);

+dfrac (sin npi )(n+dfrac {1)(n)}]

+dfrac (sin n)({2)^n} ;-|||-(2) _(n)=1+dfrac (1)({2)^2}+dfrac (1)({3)^2}+... +df

若数列_(n)=sin dfrac (n)(2)pi , 则数列_(n)=sin dfrac (n)(2)pi 发散正确错误若数列,则数列发散正确错误

证明：级数sum _(n=1)^infty (sin dfrac (1)({n)^2})收敛。证明：级数收敛。

[单选题] sum _(n=0)^infty ((-1))^ndfrac (2n+3)((2n+1)!)=A.sin1+cos1B.2sin1+cos1C.2s