an=2n-1, 设bn=1/[an*a(n+1) - 知识问答

an=2n-1, 设bn=1/[an*a(n+1)],前n项和为Tn,求证Tn<1/2

1个回答

n=1/[an*a(n+1)],
=1/(2n-1)(2n+1)
=1/2[1/(2n-1)-1/(2n+1)]
Tn=1/2[1/1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)

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