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一个简单的极限

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zhcosin Posted at 2017-5-27 22:13:52 |Read mode
Last edited by zhcosin at 2017-6-2 17:29:00题目 已知$p+1$个实数$a_i$满足$a_0+a_1+\cdots+a_p=0$,求证
\[  \lim_{n \to \infty} (a_0\sqrt{n}+a_1\sqrt{n+1}+\cdots+a_p\sqrt{n+p}) = 0  \]

证明 把$a_p$用其它$p$个数表示出来,那个式子就成为
\[ a_0(\sqrt{n}-\sqrt{n+p})+a_1(\sqrt{n+1}-\sqrt{n+p})+\cdots+a_{p-1}(\sqrt{n+p-1}-\sqrt{n+p})   \]
由此便知其极限为零。

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青青子衿 Posted at 2020-9-11 22:52:36
回复 1# zhcosin
【根式】【极限】
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