Question

设正项级数∞n＝1an收敛，证明级数∞n＝1nanan+1…a2n?1收敛

设正项级数∞n＝1an收敛，证明级数∞n＝1nanan+1…a2n?1收敛．

Answer 1

利用几何算术平均值不等式可得，
nanan+1…a2n?1=n1n(n+1)…(2n?1)?nnan(n+1)an+1…(2n?1)a2n?1
≤1nn(n+1)…(2n?1)?nan+(n+1)an+1+…+(2n?1)a2n?1n
≤1n?nan+(n+1)an+1+…+(2n?1)a2n?1n
≤2n(2n?1)n2?a1+2a2+…+(2n?1)a2n?12n(2n?1)
≤4a1+2a2+…+(2n?1)a2n?12n(2n?1)．
令bn=a1+2a2+…+nan，
由于∞
n＝1an收敛，故有∞
n＝1bnn(n+1)收敛，
从而∞
n＝1b2n?1(2n?1)(2n)收敛，
故∞
n＝1nanan+1…a2n?1收敛．