Question

高数 求数列极限

要详细过程

Answer 1

Answer 2

L = lim(n->∞) [ (2n)!/(n!.n^n) ]^[1/(n+1)]
lnL
=lim(n->∞) [1/(n+1) ] { ∑(i:1->n) ln[(n+i)/n ] }
=lim(n->∞) [1/(n+1) ] { ∑(i:1->n) ln[(1 + i/n) }
=∫(0->1) ln(1+x) dx
=[ xln(1+x) ]|(0->1) -∫(0->1) x/(1+x) dx
=ln2 - ∫(0->1) [1- 1/(1+x)] dx
=ln2 - [ x- ln|1+x| ]|(0->1)
=ln2- ( 1- ln2)
=2ln2 -1
L = e^(2ln2-1)= 4e^(-1)
lim(n->∞) [ (2n)!/(n!.n^n) ]^[1/(n+1)] = 4e^(-1)

追问

请问第二步是怎么得出的啊

追答

L
= lim(n->∞) [ (2n)!/(n!.n^n) ]^[1/(n+1)]
=lim(n->∞) [ (n+1)(n+2)....(2n)/(n^n) ]^[1/(n+1)]
=lim(n->∞){ [ (n+1)/n] [(n+2)/n].....[ (n+n)/n] }^[1/(n+1)]
lnL
=lim(n->∞) [1/(n+1) ] { ∑(i:1->n) ln[(n+i)/n ] }