怎么用定义判断级数∑sin(nπ/6)的敛散性？
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简单分析一下，答案如图所示
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解：由于一般项
u[n]=sin(nπ/6)
=[2sin(π/12)sin(nπ/6)]/[2sin(π/12)]
=[cos((2n-1)π/12)-cos((2n+1)π/12)]
/[2sin(π/12)]，
从而部分和
s[n]
=[cos(π/12)-cos(3π/12)
+cos(3π/12)-cos(5π/12)
+cos(5π/12)-cos(7π/12)
+…
…
+cos((2n-1)π/12)-cos((2n+1)π/12)]
/[2sin(π/12)]
=[cos(π/12)-cos((2n+1)π/12)]/[2sin(π/12)]，
因为当n→∞时，cos((2n+1)π/12)的极限不存在，所以s[n]的极限不存在，所以级数
Σsin(nπ/6)发散.
u[n]=sin(nπ/6)
=[2sin(π/12)sin(nπ/6)]/[2sin(π/12)]
=[cos((2n-1)π/12)-cos((2n+1)π/12)]
/[2sin(π/12)]，
从而部分和
s[n]
=[cos(π/12)-cos(3π/12)
+cos(3π/12)-cos(5π/12)
+cos(5π/12)-cos(7π/12)
+…
…
+cos((2n-1)π/12)-cos((2n+1)π/12)]
/[2sin(π/12)]
=[cos(π/12)-cos((2n+1)π/12)]/[2sin(π/12)]，
因为当n→∞时，cos((2n+1)π/12)的极限不存在，所以s[n]的极限不存在，所以级数
Σsin(nπ/6)发散.
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