lim(n→无穷)[1/(n^2+1)+2/(n^2+2)+.+2n/(n^2+n)]
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lim(n→无穷)[1/(n^2+1)+2/(n^2+2)+.+2n/(n^2+n)]lim(n→无穷)[1/(n^2+1)+2/(n^2+2)+.+2n/(n^2+n)]lim(n→无穷)[1/(n
lim(n→无穷)[1/(n^2+1)+2/(n^2+2)+.+2n/(n^2+n)]
lim(n→无穷)[1/(n^2+1)+2/(n^2+2)+.+2n/(n^2+n)]
lim(n→无穷)[1/(n^2+1)+2/(n^2+2)+.+2n/(n^2+n)]
1/(n^2+1)+2/(n^2+2)+.+n/(n^2+n)
≤1/(n^2+1)+2/(n^2+1)+.+n/(n^2+1)
≤(n^2+n)/2(n^2+1)····（1）
1/(n^2+1)+2/(n^2+2)+.+n/(n^2+n)
≥1/(n^2+n)+2/(n^2+n)+.+n/(n^2+n)
≥(n^2+n)/2(n^2+n)
=1/2
(1)的极限为1/2 根据夹逼定理 极限为1/2
...这个题目还有点小问题，最后一项应该是n/(n^2+n) 或者是2n/(n^2+2n)
但是不影响过程
以最后一项是n/(n^2+n)为例
方法是使用夹逼定理
对任意1<=i<=n i/(n^2+n) <=i/(n^2+i)<=i/n^2
lim(1+2+...n)/(n^2+n)<= 原式<=lim (1+2+...n)/n^2
即...
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...这个题目还有点小问题，最后一项应该是n/(n^2+n) 或者是2n/(n^2+2n)
但是不影响过程
以最后一项是n/(n^2+n)为例
方法是使用夹逼定理
对任意1<=i<=n i/(n^2+n) <=i/(n^2+i)<=i/n^2
lim(1+2+...n)/(n^2+n)<= 原式<=lim (1+2+...n)/n^2
即 lim (n^2+n)/2(n^2+n)<=原式<=lim (n^2+n)/2n^2
即 1/2<=原式<=1/2
所以 原式=1/2
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