S=(1/3^2-1)+(1/4^2-1).....+(1/n^2-1)+1/(n+1)^2-1,求和。。
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S=(1/3^2-1)+(1/4^2-1).....+(1/n^2-1)+1/(n+1)^2-1
=1/(3-1)(3+1)+1/(4-1)(4+1)+.....+1/(n-1)(n+1)+1/(n+1-1)(n+1+1)
=1/2[1/2-1/4+1/3-1/5+....+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2{[(1/2+1/3+....1/(n-1)+1/n]-[(1/4+1/5+....+1/(n+1)+1/(n+2)]}
=1/2[1/2+1/3-1/(n+1)-1/(n+2)]
=5/12-1/2[1/(n+1)+1/(n+2)]
当n趋向于无穷时，S=5/12
=1/(3-1)(3+1)+1/(4-1)(4+1)+.....+1/(n-1)(n+1)+1/(n+1-1)(n+1+1)
=1/2[1/2-1/4+1/3-1/5+....+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2{[(1/2+1/3+....1/(n-1)+1/n]-[(1/4+1/5+....+1/(n+1)+1/(n+2)]}
=1/2[1/2+1/3-1/(n+1)-1/(n+2)]
=5/12-1/2[1/(n+1)+1/(n+2)]
当n趋向于无穷时，S=5/12
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