Question

高数这道极限题怎么写？

Answer 1

因为∑(k=1->n) (lnk)^2

=∑(k=1->n) (lnk-lnn+lnn)^2
=∑(k=1->n) [(lnk-lnn)^2+2(lnk-lnn)lnn+(lnn)^2]
=∑(k=1->n) (lnk-lnn)^2+∑(k=1->n) 2(lnk-lnn)lnn+∑(k=1->n) (lnn)^2
=∑(k=1->n) [ln(k/n)]^2+2lnn*∑(k=1->n) ln(k/n)+n*(lnn)^2
所以(1/n)*∑(k=1->n) (lnk)^2
=(1/n)*∑(k=1->n) [ln(k/n)]^2+2lnn*(1/n)*∑(k=1->n) ln(k/n)+(lnn)^2
又因为∑(k=1->n) lnk
=∑(k=1->n) (lnk-lnn+lnn)
=∑(k=1->n) (lnk-lnn)+∑(k=1->n) lnn
=∑(k=1->n) ln(k/n)+n*lnn
所以[(1/n)*∑(k=1->n) lnk]^2
=[(1/n)*∑(k=1->n) ln(k/n)+lnn]^2
=[(1/n)*∑(k=1->n) ln(k/n)]^2+2lnn*(1/n)*∑(k=1->n) ln(k/n)+(lnn)^2
原式=lim(n->∞) {(1/n)*∑(k=1->n) [ln(k/n)]^2+2lnn*(1/n)*∑(k=1->n) ln(k/n)+(lnn)^2-[(1/n)*∑(k=1->n) ln(k/n)]^2-2lnn*(1/n)*∑(k=1->n) ln(k/n)-(lnn)^2}
=lim(n->∞) {(1/n)*∑(k=1->n) [ln(k/n)]^2-[(1/n)*∑(k=1->n) ln(k/n)]^2}
=∫(0,1) (lnx)^2dx-[∫(0,1) lnxdx]^2
=x(lnx)^2|(0,1)-∫(0,1) xd[(lnx)^2]-[(xlnx-x)|(0,1)]^2
=-∫(0,1) 2lnxdx-1
=-2(xlnx-x)|(0,1)-1
=1