100很大,你可以设通项为(n^3-1)/(n^3+1),原式为n从2到100通项的乘积
∏(n^3-1)/(n^3+1)
=∏(n-1)[n(n+1)+1]/{(n+1)[(n-1)n+1]}
=∏[(n-1)/(n+1)]*∏{[n(n+1)+1]/[(n-1)n+1]}
={2/[n(n+1)]}{[n(n+1)+1]/(1*2+1)}
=(2/3){[n(n+1)+1]/[n(n+1)]}
[n(n+1)+1]/[n(n+1)]当n很大时,趋于1
∴(2/3){[n(n+1)+1]/[n(n+1)]}趋于2/3
最接近的话 ,应该是1