Question

大一高等数学A题、1/( n^2+1)+2/(n^2+2)+……+n/(n^2+n) 的极限怎么求~

Answer 1

因为n^2+1<n^2+2<......<n^2+n,所以原式>=1/（n^+n)+2/（n^+n)+3/（n^+n)+........+n/（n^+n)=(1+2+.......+n)/（n^+n)=1/2.同时原式<=1/(n^2+1)+2/(n^2+1)+3/(n^2+1)+........+n/(n^2+1)=(1+1/n)/2.而（1+1/n)/2）的极限是1/2。所以原式的极限=1/2。

Answer 2

设[1/(n^2+1)+2/(n^2+2)+....+n/(n^2+n)]=S
则S<[1/(n^2+1)+2/(n^2+1)+....+n/(n^2+1)]=(1+2+……+n）/(n^2+1)=n*(n+1)/2(n^2+1)
因为S>[1/(n^2+n)+2/(n^2+n)+....+n/(n^2+1)]=n*(n+1)/2(n^2+n)
且limn*(n+1)/2(n^2+1)=1/2(n趋于无穷)
limn*(n+1)/2(n^2+n)=1/2(n趋于无穷)
lim[1/(n^2+1)+2/(n^2+2)+....+n/(n^2+n)] n趋于无穷=1/2

Answer 3

lim（n→∞）[1/（n^2+n+1）+2/（n^2+n+2）+3/（n^2+n+3）+……+n/（n^2+n+n）]＝ lim（n→∞）[(1+2+...+n)/（n^2+n+1）]=lim（n→∞）1/2*[n(n+1)/（n^2+n+1）]=1/2