两道级数收敛题,求大神指导
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两道级数收敛题,求大神指导两道级数收敛题,求大神指导两道级数收敛题,求大神指导(3)取a[n]=1/n,b[n]=sin(n)·sin(n²).有2b[n]=2sin(n)·sin(n
两道级数收敛题,求大神指导
两道级数收敛题,求大神指导
两道级数收敛题,求大神指导
(3) 取a[n] = 1/n,b[n] = sin(n)·sin(n²).
有2b[n] = 2sin(n)·sin(n²) = cos(n²-n)-cos(n+n²) = cos((n-1)n)-cos(n(n+1)).
部分和∑{m < k ≤ n} b[k] = (cos(m(m+1))-cos(n(n+1)))/2绝对值不大于1,即有界.
另一方面,易见a[n]单调趋于0.
根据Dirichlet判别法,级数∑sin(n)·sin(n²)/n = ∑a[n]·b[n]收敛.
(4) 取a[n] = 1/n,b[n] = sin(n).
有2sin(1/2)·b[n] = 2sin(1/2)sin(n) = cos(n-1/2)-cos(n+1/2).
部分和∑{m < k ≤ n} b[k] = (cos(m+1/2)-cos(n+1/2))/(2sin(1/2))绝对值不大于1/(2sin(1/2)),即有界.
另一方面,易见a[n]单调趋于0.
根据Dirichlet判别法,级数∑sin(n)/n = ∑a[n]·b[n]收敛.
由|sin(n+1/n)-sin(n)| = |2cos(n+1/(2n))sin(1/(2n))| ≤ 2sin(1/(2n)) < 1/n.
于是0 ≤ |sin(n+1/n)-sin(n)|/n < 1/n².
根据比较判别法,可知级数∑|sin(n+1/n)-sin(n)|/n收敛.
即∑(sin(n+1/n)-sin(n))/n绝对收敛,从而也是收敛的.
因此∑sin(n+1/n)/n作为两个收敛级数的和是收敛的.