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幼苗
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(1+1/a1)(1+1/a2).(1+1/an)
=(1+1/1)(1+1/3)(1+1/5).(1+1/(2n-1))
=(2/1)(4/3)(6/5)...[(2n)/(2n-1)].
设b(n)=[(1+1/a1)(1+1/a2)...(1+1/an)]/√(2n+1)
=(2/1)(4/3)(6/5)...[(2n)/(2n-1)]/√(2n+1).
b(n+1)/b(n)=(2n+2)/[√(2n+1)√(2n+3)]>1.
2√3/3=b(1)
1年前
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