用Cauchy收敛原理证明下面数列收敛xn=sin2x2(2+sin2x)+sin3x3(3+sin3x)+…+sinnxn(n+sinnx)．

Question

用Cauchy收敛原理证明下面数列收敛x_n=(sin2x)/(2(2+sin2x))+(sin3x)/(3(3+sin3x))+…+(sinnx)/(n(n+sinnx))．

Answer 1

由题意，对任意的自然数n，和正整数p，有
|x_n+p-x_n|=|(sin(n+1)x)/((n+1)(n+1+sin(n+1)x))+(sin(n+2)x)/((n+2)(n+2+sin(n+2)x))+…+(sin(n+p)x)/((n+p)(n+p+sin(n+p)x))|
≤(1)/(n(n+1))+(1)/((n+1)(n+2))+…+(1)/((n+p-1)(n+p))
=((1)/(n)-(1)/(n+1))+((1)/(n+1)-(1)/(n+2))+…+((1)/(n+p-1)-(1)/(n+p))
=(1)/(n)-(1)/(n+p)<(1)/(n)
任取ɛ>0，取自然数N=[(1)/(ɛ)]，则对任意的整数n>N，以及正整数p，均有|xn-xn+p|<(1)/(n)≤ɛ成立，
因此数列{x_n}收敛．