利用极限存在准则证明:limn趋向于无穷,n【1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)】=1
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![利用极限存在准则证明:limn趋向于无穷,n【1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)】=1](/uploads/image/z/6795602-26-2.jpg?t=%E5%88%A9%E7%94%A8%E6%9E%81%E9%99%90%E5%AD%98%E5%9C%A8%E5%87%86%E5%88%99%E8%AF%81%E6%98%8E%EF%BC%9Alimn%E8%B6%8B%E5%90%91%E4%BA%8E%E6%97%A0%E7%A9%B7%2Cn%E3%80%901%2F%EF%BC%88n%5E2%2B%CF%80%EF%BC%89%2B1%2F%EF%BC%88n%5E2%2B2%CF%80%EF%BC%89%2B...%2B1%2F%EF%BC%88n%5E2%2Bn%CF%80%29%E3%80%91%3D1)
利用极限存在准则证明:limn趋向于无穷,n【1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)】=1
利用极限存在准则证明:limn趋向于无穷,n【1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)】=1
利用极限存在准则证明:limn趋向于无穷,n【1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)】=1
证明: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.
lim n【1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)】
=lim 1/(n+π/n)+1/(n+2π/n)+...+1/(n+π)】
=lim n*1/n
=1
迫敛准则
设 u(n) =n【1/(n^2+π)+1/(n^2+2π)+ ... +1/(n^2+nπ)】
n * n /(n^2+nπ) < u(n) < n * n / (n^2+π)
lim n->∞ n^2 /(n^2+nπ) = lim n->∞ n^2 / (n^2+π) = 1
lim n->∞ u(n)=1
夹逼准则n^2/(n^2+nπ)>n【1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)】>n^2/(n^2+π)
n^2/(n^2+nπ)=n^2/(n^2+π)=1(当n趋向∞)