1) lim(1-1/2^2)(1-1/3^2)...(1-1/n^2)
首先,我們發(fā)現(xiàn),1-(1/n)^2=(n-1)/n*(n+1)/n
所以,(1-1/2^2)(1-1/3^2)...(1-1/n^2)=1/2*3/2*2/3*4/3*...*(n-1)/n*(n+1)/n=(n+1)/(2n).
因此,原極限為lim (n+1)/(2n) = 1/2
2)limn^2[(k/n)-(1/n+1)-(1/n+2)-...-1/(n+k)]
首先,我們發(fā)現(xiàn),1/n-1/(n+k)=k/n(n+k)
所以,(k/n)-(1/n+1)-(1/n+2)-...-1/(n+k)=1/n(n+1)+2/n(n+2)+...+k/n(n+k)
因此,原極限為lim n^2*[1/n(n+1)+2/n(n+2)+...+k/n(n+k)]=1+2+...+k=k(k+1)/2