已知an=n^2,求证1⼀a1+1⼀a2+…+1⼀an<7⼀4

2025-01-02 12:21:55
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回答1:

解:原式=1/1+1/4+1/9+1/16+。。。。。。。+1/N^2
原式<式(1)=1+1/(1*3)+1/(2*4)+1/(3*5).........+1/[(n-1)(n+1)]
=1+1/2[1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+.......+1/(n-1)-1/(n+1) ]
=1+1/2{1+1/2-[1/(n+1)+1/n]}
=1+1/2*3/2-1/2[1/(n+1)+1/n]
=7/4-1/2[1/(n+1)+1/n]<7/4