Question

cosnπ/n是否收敛

无穷和Σcosn/n是否收敛?
Σcosn是不收敛的,但是它类似交错级数,而且正负的机会对半
Σ1/n也是不收敛的,但是它处在收敛和不收敛的边缘
那么Σcosn/n会不会收敛呢?它的波动越来越小,而且正负相抵

Answer 1

有没有学过Dirichlet判别法?
如果数列a[n]单调趋于0, 同时级数∑b[n]的部分和有界, 则级数∑a[n]·b[n]收敛.
取a[n] = 1/n, 易见其单调趋于0.
取b[n] = cos(n), 有
b[1]+b[2]+...+b[n] = cos(1)+cos(2)+...+cos(n)
= (2cos(1)sin(1/2)+2cos(2)sin(1/2)+...+2cos(n)sin(1/2))/(2sin(1/2))
= (sin(3/2)-sin(1/2)+sin(5/2)-sin(3/2)+...+sin(n+1/2)-sin(n-1/2))/(2sin(1/2))
= (sin(n+1/2)-sin(1/2))/(2sin(1/2)),
可知|b[1]+b[2]+...+b[n]| ≤ 1/sin(1/2), 故∑b[n]的部分和有界.
根据Dirichlet判别法, 级数∑a[n]·b[n] = ∑cos(n)/n收敛.