利用极限的夹副准则证明limn→无穷大(n⼀n^2+π+n⼀n^2+2π+...+n⼀n^2+nπ)=1

n/(n^2+nπ) ≤ n/(n^2+mπ) ≤ n/(n^2 + π) 注：n ≤ m ≤ 1
所以，
n*[n/(n^2+nπ)]=n^2/(n^2+nπ) ≤ ∑n/(n^2+mπ) ≤ n*[n/(n^2+π) = n^2/(n^2+π)

因为：lim[n^2/(n^2+nπ)]=lim[1/(1+π/n)] = 1
lim[n^2/(n^2+π)] = lim[1/(1+π/n^2)] = 1
所以，
lim∑n/(n^2+mπ) = 1

参考书本例题就可以啦

证明：limn【1/（n^2+π）+1/（n^2+2π）+...+1/（n^2+nπ)】=limn*n/n^2=limn^2/n^2=1
又因为limn【1/（n^2+π）+1/（n^2+2π）+...+1/（n^2+nπ)】>limn【(1/n^2+nπ)+(1/n^2+nπ)+......(1/n^2+nπ)】
=limn(n/(n^2+nπ)
=limn/n+π)
=1
所以limn【1/（n^2+π）+1/（n^2+2π）+...+1/（n^2+nπ)】=1 成立。

n^2/n^2+π<这个式子（简称Y吧）由于n趋于正无穷
所以 左边= 1/(1+π/n^2)=1
右边=1/(1+π/n)=1
1所以Y=1