Question

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

Answer 1

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