Question

数列极限的问题

lim[(1/1*4)+(1/4*7)+(1/7*10)+.....+1/(3n-2)(3n+2)] 求解 要有步骤

Answer 1

lim[(1/1*4)+(1/4*7)+(1/7*10)+.....+1/(3n-2)(3n+2)]
=(1/3)lim{(1 - 1/4)+(1/4 - 1/7) + ..... +[1/(3n-2) - 1/(3n+2)]}
=(1/3)lim{1 - 1/(3n+2)}
=(1/3)lim(3n+1)/(3n+2)
=(1/3)lim[3+(1/n)]/[3+(2/n)] [分子分母同除以n]
=(1/3)[3+0]/[3+0]
=1/3

or
lim[(1/1*4)+(1/4*7)+(1/7*10)+.....+1/(3n-2)(3n+2)]
=(1/3)lim{(1 - 1/4)+(1/4 - 1/7) + ..... +[1/(3n-2) - 1/(3n+2)]}
=(1/3)lim{1-1/(3n+2)}
=(1/3)[1-1/∞]
=(1/3)[1-0]
=1/3

Answer 2

拆开算,用partial fraction decomposition,不就解决了??