试证:对任意正整数n,有1/(1*2*3)+1/(2*3*4)+…+1/(n(n+1)(n+2))

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试证:对任意正整数n,有1/(1*2*3)+1/(2*3*4)+…+1/(n(n+1)(n+2))

试证:对任意正整数n,有1/(1*2*3)+1/(2*3*4)+…+1/(n(n+1)(n+2))
试证:对任意正整数n,有1/(1*2*3)+1/(2*3*4)+…+1/(n(n+1)(n+2))

试证:对任意正整数n,有1/(1*2*3)+1/(2*3*4)+…+1/(n(n+1)(n+2))
因为1/(1*2*3)=(1/2)*[1/(1*2)-1/(2*3)],
1/(2*3*4)=(1/2)*[1/(2*3)-1/(3*4)],
...
(可以把右边通分,证明等式成立)
所以1/(1*2*3)+1/(2*3*4)+...+1/n(n+1)(n+2)
=(1/2)*[1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+...+1/n(n+1)-1/(n+1)(n+2)]
=(1/2)*[1/2-1/(n+1)(n+2)]
=1/4-1/2(n+1)(n+2)
因为1/2(n+1)(n+2)>0,所以式子左边=1/4-1/2(n+1)(n+2)

原来的通项公式为1/(n(n+1)(n+2))=1/2(1/n(n+1)-1/(n+1)(n+2))
相加得Sn=1/2(1/2-1/(n+1)(n+2))恒小于1/4 ,(括号里小于1/2)