【急】求数学大神证明(n趋于无穷大)lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/03 06:04:33

【急】求数学大神证明(n趋于无穷大)lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))=1用夹逼准则证

a≥0时,有
lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))≥lim n(1/(n^2+a)+1/(n^2+a)+...+1/(n^2+a))
=lim n(n/(n^2+a))=lim(n^2/(n^2+a))=lim(1/(1+a/n^2))=1 (n->∞)
同时有
lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))≤lim n(1/(n^2+na)+1/(n^2+na)+...+1/(n^2+na))
=lim n(n/(n^2+na))=lim(n^2/(n^2+na))=lim(1/(1+a/n))=1 (n->∞)
即同时有 1≤lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))≤1
∴lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))=1
a≤0时,有
lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))≤lim n(1/(n^2+na)+1/(n^2+na)+...+1/(n^2+na))
=lim n(n/(n^2+na))=lim(n^2/(n^2+na))=lim(1/(1+a/n))=1 (n->∞)
同时有
lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))≥lim n(1/(n^2+a)+1/(n^2+a)+...+1/(n^2+a))
=lim n(n/(n^2+a))=lim(n^2/(n^2+a))=lim(1/(1+a/n^2))=1 (n->∞)
即同时有 1≤lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))≤1
∴lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))=1
综上所述,lim n(1/(n^2+a)+1/(n^2+2a)+...+1/(n^2+na))=1

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