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5656

2024-07-02 20:25:01   216人阅读

$\bf命题1：$设$f(x)$是$\left[ {1, + \infty } \right)$上的非负单调减少函数，令

an=∑k=1nf(k)?∫n1f(x)dx,n∈N+
<script id="MathJax-Element-1" type="math/tex; mode=display">{a_n} = \sum\limits_{k = 1}^n {f\left( k \right)} - \int_1^n {f\left( x \right)dx} ,n \in {N_ + }</script>
证明：数列$\left\{ {{a_n}} \right\}$收敛

证明：由$f(x)$在$\left[ {1, + \infty } \right)$上单调减少知，$f(x)$在$\left[ {n,n + 1} \right]$上可积，且

f(n+1)≤∫n+1nf(x)dx≤f(n),n∈N+
<script id="MathJax-Element-2" type="math/tex; mode=display">\begin{equation}\label{eq1}f\left( {n + 1} \right) \le \int_n^{n + 1} {f\left( x \right)dx} \le f\left( n \right),n \in {N_ + }\end{equation}</script>

从而可知

an+1?an=f(n+1)?∫n+1nf(x)dx≤0
<script id="MathJax-Element-3" type="math/tex; mode=display">\begin{equation}\label{eq2}{a_{n + 1}} - {a_n} = f\left( {n + 1} \right) - \int_n^{n + 1} {f\left( x \right)dx} \le 0\end{equation}</script>
即$\left\{ {{a_n}} \right\}$单调减少；而又由$\eqref {eq1}$知

an=∑k=1nf(k)?∫n1f(x)dx≥∑k=1n∫k+1kf(x)dx?∫n1f(x)dx=∫n+1nf(x)dx≥f(n+1)≥0 
<script id="MathJax-Element-4" type="math/tex; mode=display">\begin{align}\label{eq3} {a_n} &= \sum\limits_{k = 1}^n {f\left( k \right)} - \int_1^n {f\left( x \right)dx} \nonumber\\& \ge \sum\limits_{k = 1}^n {\int_k^{k + 1} {f\left( x \right)dx} } - \int_1^n {f\left( x \right)dx}\\& = \int_n^{n + 1} {f\left( x \right)dx} \ge f\left( {n + 1} \right) \ge 0 \nonumber\end{align}</script>
即$\left\{ {{a_n}} \right\}$有上界，故由$\eqref {eq2}$,$\eqref {eq3}$及单调有界原理知数列$\left\{ {{a_n}} \right\}$收敛


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