Question

求lim(n→∞) n(1/(n^2+1)+1/(n^2+1)+...1/(n^2+n)

Answer 1

题目是lim(n→∞) n(1/(n^2+1)+1/(n^2+2)+...1/(n^2+n)吧？
这个用夹逼定理
lim(n→∞) n(1/(n^2+n)+1/(n^2+n)+...1/(n^2+n)≤lim(n→∞) n(1/(n^2+1)+1/(n^2+2)+...1/(n^2+n)≤
lim(n→∞) n(1/(n^2+1)+1/(n^2+1)+...1/(n^2+1)

lim(n→∞) n^2/(n^2+n)≤lim(n→∞) n(1/(n^2+1)+1/(n^2+2)+...1/(n^2+n)≤
lim(n→∞) n^2/(n^2+1)
1≤lim(n→∞) n(1/(n^2+1)+1/(n^2+2)+...1/(n^2+n)≤1
所以lim(n→∞) n(1/(n^2+1)+1/(n^2+2)+...1/(n^2+n)=1

Answer 2

1/(n^2+n)<1/(n^2+k)<1/n^2,1<=k<=n，对k求和得
n/(n^2+n)<关于k的和<n/n^2，
于是n^2/(n^2+n)<待求表达式<n^2/n^2，夹逼定理知道极限是1。

Answer 3

lim(n→∞) n(1/(n^2+1)+1/(n^2+2)+...1/(n^2+n) )
=lim(n→∞) (1/(1+1/n^2)+1/(1+2/n^2)+...+ 1/(1+1/n) )
=∞