lim[1/1*2*3+1/2*3*4+…+1/n(n+1)(n+2)] ,n趋向无穷大的时候,极限多少,怎么算的
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1/n(n+1)(n+2)
= 1/(n+1) {（1/2）[ 1/n - 1/(n+2) ]}
=（1/2）[ 1/(n+1)1/n - 1/(n+1)1/(n+2) ]
=（1/2）[ (1/n - 1/(n+1)) - (1/(n+1)-1/(n+2))]
=（1/2）[ 1/n - 2/(n+1)) +1/(n+2) ]
1/1*2*3+1/2*3*4+…+1/n(n+1)(n+2)
=1/2 { [1 - 2/2 +1/3] +[1/2 -2/3+1/4]+ ...+[ 1/n - 2/(n+1)) +1/(n+2) ]}
=1/2 { [1+1/2+1/3+...+1/n] - 2[1/2+1/3+...+1/(n+1)] +[1/3+1/4+...+1/(n+2) ]}
=1/2 { [1-1/2-1/(n+1)+1/(n+2) ]}
lim[1/1*2*3+1/2*3*4+…+1/n(n+1)(n+2)]
=lim 1/2*{ [1-1/2-1/(n+1)+1/(n+2) ]}
=1/4
= 1/(n+1) {（1/2）[ 1/n - 1/(n+2) ]}
=（1/2）[ 1/(n+1)1/n - 1/(n+1)1/(n+2) ]
=（1/2）[ (1/n - 1/(n+1)) - (1/(n+1)-1/(n+2))]
=（1/2）[ 1/n - 2/(n+1)) +1/(n+2) ]
1/1*2*3+1/2*3*4+…+1/n(n+1)(n+2)
=1/2 { [1 - 2/2 +1/3] +[1/2 -2/3+1/4]+ ...+[ 1/n - 2/(n+1)) +1/(n+2) ]}
=1/2 { [1+1/2+1/3+...+1/n] - 2[1/2+1/3+...+1/(n+1)] +[1/3+1/4+...+1/(n+2) ]}
=1/2 { [1-1/2-1/(n+1)+1/(n+2) ]}
lim[1/1*2*3+1/2*3*4+…+1/n(n+1)(n+2)]
=lim 1/2*{ [1-1/2-1/(n+1)+1/(n+2) ]}
=1/4
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