试证；对任意的正整数n，有1/(1*2*3)+1/(2*3*4)+.....+1/n(n+1)(n+2)<1/4
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1/n(n+1)(n+2)=1/n·[1/(n+1) - 1/(n+2)]=1/n(n+1) - 1/n(n+2) =[1/n-1/(n+1)] - ½[1/n-1-(n+2)]
=½[1/n-2/(n+1)+1/(n+2)].
∴原式=½(1/1-2/2+1/3)+½(1/2-2/3+1/4)+½(1/3-2/4+1/5)+···+½[1/n-2/(n+1)+1/(n+2)]
=½(1/1+1/2+1/3+···+1/n) - 1·[1/2+1/3+1/4+···+1/(n+1)] + ½[1/3+1/4+···+1/(n+2)]
=½[1+1/2+1/(n+1)+1/(n+2)] + 1·[1/3+1/4+···+1/n] - 1·[1/2+1/3+1/4+···+1/(n+1)]
=1/2+1/4 + ½[1/(n+1)+1/(n+2)] - [1/2+1/(n+1)]
=1/4-½[1/(n+1)-1/(n+2)]
<1/4.
=½[1/n-2/(n+1)+1/(n+2)].
∴原式=½(1/1-2/2+1/3)+½(1/2-2/3+1/4)+½(1/3-2/4+1/5)+···+½[1/n-2/(n+1)+1/(n+2)]
=½(1/1+1/2+1/3+···+1/n) - 1·[1/2+1/3+1/4+···+1/(n+1)] + ½[1/3+1/4+···+1/(n+2)]
=½[1+1/2+1/(n+1)+1/(n+2)] + 1·[1/3+1/4+···+1/n] - 1·[1/2+1/3+1/4+···+1/(n+1)]
=1/2+1/4 + ½[1/(n+1)+1/(n+2)] - [1/2+1/(n+1)]
=1/4-½[1/(n+1)-1/(n+2)]
<1/4.
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x小于1/4吧
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