证明:∑{1≤n}[sin(n^2)sin(n)]/n收敛.
T(n)=∑{1≤k≤n}[sin(k^2)sin(k)]=
=(1/2)∑{1≤k≤n}[cos(k(k-1))-cos((k+1)k)]=
=[1-cos(n(n+1))]/2
==>
|T(n)|≤1
m≤n,
R(m,n)=∑{m≤k≤n}[sin(k^2)sin(k)]/k=
=∑{m≤k≤n}[(T(k)-T(k-1))/k]=
=-T(m-1)/m+∑{m≤k≤n-1}{T(k)/[(k+1)k]}+T(n)/n=
==>
|R(m,n)|≤1/m+∑{m≤k≤n-1}{1/[(k+1)k]}+1/n=2/m
==>
任意ε>0,取N>0,使2/N<ε,
当任意N≤m≤n时,
|R(m,n)|≤2/m≤2/N<ε,
所以∑{1≤n}[sin(n^2)sin(n)]/n收敛.
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