+dfrac (1)((2n-1)(2n+1))]

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

+dfrac (1)({(2n-1))^2}(2n-1)x+... ] -|||-(B) dfrac (2)(pi )[ dfrac (1)({2)^2}sin

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

(2) lim _(narrow infty )dfrac (3n+1)(2n+1)=dfrac (3)(2) ;

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

下列级数绝对收敛的是()A.sum _(n=1)^infty dfrac ({(-1))^n+1}(2n+1)B.sum _(n=1)^infty dfrac

+(2n-1))({n)^2}=求极限

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