北京师范大学1999分析一题

记$\displaystyle C_0:=\lim\limits_{n \rightarrow +\infty}\left\{\sum\limits_{k=1}^n\frac{1}{k}-\ln (n+1) \right\}$, 试证明: $\displaystyle \lim\limits_{p \rightarrow 0^+}\left(\sum\limits_{n=1}^{+\infty}\frac{1}{n^{1+p}}-\frac{1}{p} \right)=C_0.$
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妖心儿

赞同来自: Math001 行如海 JiYinX dingding 许高谦

$n\le x\le n+1$时,
$0\le\cfrac 1{n^{1+p}}-\cfrac 1{x^{1+p}}\\
=-(1+p)\xi^{-(2+p)}(n-x)\\
=\cfrac{1+p}{\xi^{2+p}}(x-n)\\
\le\cfrac {1+p}{n^{2+p}}(x-n)\\
\le\cfrac 2{n^2}(x-n)$

对x在[n,n+1]积分,$0\le\cfrac 1{n^{1+p}}-\int_n^{n+1}\cfrac 1{x^{1+p}}dx\le\cfrac 1{n^2}\\
\Rightarrow 0\le\sum\limits_{n=N+1}^{\infty}\cfrac 1{n^{1+p}}-\int_{N+1}^{\infty}\cfrac 1{x^{1+p}}dx\le\cfrac 1N$

$|\sum\limits_{n=1}^{\infty}\cfrac 1{n^{1+p}}-\cfrac 1p-C_0|=|\sum\limits_{n=1}^{\infty}\cfrac 1{n^{1+p}}-\int_{1}^{\infty}\cfrac 1{x^{1+p}}dx-C_0|\\
\le|\sum\limits_{n=1}^{N}\cfrac 1{n^{1+p}}-\int_{1}^{N+1}\cfrac 1{x^{1+p}}dx-C_0|+|\sum\limits_{n=N+1}^{\infty}\cfrac 1{n^{1+p}}-\int_{N+1}^{\infty}\cfrac 1{x^{1+p}}dx|\\
\le|\sum\limits_{n=1}^{N+1}\cfrac 1{n^{1+p}}-\int_{1}^{N}\cfrac 1{x^{1+p}}dx-C_0|+\cfrac 1N$

$\limsup\limits_{p\to 0^+}|\sum\limits_{n=1}^{\infty}\cfrac 1{n^{1+p}}-\cfrac 1p-C_0|\\
\le\limsup\limits_{p\to 0^+}|\sum\limits_{n=1}^{N}\cfrac 1{n^{1+p}}-\int_{1}^{N+1}\cfrac 1{x^{1+p}}dx-C_0|+\cfrac 1N\\
=|\sum\limits_{n=1}^{N}\cfrac 1{n}-\int_{1}^{N+1}\cfrac 1{x}dx-C_0|+\cfrac 1N\\
=|\sum\limits_{n=1}^{N}\cfrac 1{n}-ln(N+1)-C_0|+\cfrac 1N$

令$N\to\infty$得$\limsup\limits_{p\to 0^+}|\sum\limits_{n=1}^{\infty}\cfrac 1{n^{1+p}}-\cfrac 1p-C_0|=0$
即$\lim\limits_{p\to 0^+} \sum\limits_{n=1}^{\infty}\cfrac 1{n^{1+p}}-\cfrac 1p=C_0$.

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