+dfrac (1)({(2n-1))^2}(2n-1)x+... ] -|||-(B) dfrac (2)(pi )[ dfrac (1)({2)^2}sin 2x+dfrac (1)({4)^2}sin 4x+dfrac (1)({6)^2}sin 6x+... +dfrac (1)({(2n))^2}sin 2nx+... ] -|||-(C) dfrac (4)(n)[ cos x+dfrac (1)({3)^2}cos 3x+dfrac (1)({5)^2}cos x+... +dfrac (1)({(2n-1))^2}cos (2n-1)x+... ] -|||-(D) dfrac (1)(pi )[ dfrac (1)({2)^2}cos 2x+dfrac (1)({4)^2}cos 4x+dfrac (1)({6)^2}cos 6x+... +dfrac (1)({(2n))^2}cos 2nx+... ]

+dfrac (1)(n(n+1)) =-|||-(3) lim _(narrow infty )(dfrac (1)(2)+dfrac (3)({2)^2}+

,求幂级数 sum _(n=1)^infty dfrac (2n-1)({2)^n}(x)^2n-2 的和函数,并求级数 sum _(n=1)^infty df

+dfrac (1)((2n-1)(2n+1))]求极限

幂级数sum _(n=1)^infty dfrac ({(-1))^n}(2n-1)(x)^2n-1(|x|lt 1)的和函数sum _(n=1)^infty

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

5.已知 (x)=sum _(n=1)^infty ((-1))^n-1dfrac (1)((2n-1)!)((pi x))^2n-1 则 f(1)=()-||

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

(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}

(B) dfrac (1)(2)(X)_(1)+dfrac (1)(2)(X)_(2)-|||-(C) dfrac (1)(2)(X)_(1)+dfrac (1

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}((cos )^2x+dfrac (xcos x)(1+{cos )^2x})dx