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

Question

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

Answer 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