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题：试证\(\;{\displaystyle\lim_{x\to\infty}}\big({\large\frac{\Gamma'(x)}{\Gamma(x)}}-\ln x\big)=0.\)w
证：\(\;\because\;\Gamma(x+1)=x\Gamma(x),\;\ln\Gamma(x+1)-\ln\Gamma(x)=\ln x.\,\)hgL
\((^*)\quad\exists s\in(x,x+1):\,{(\ln\Gamma(z))'\big|_{z=s}=\large\frac{\Gamma'(s)}{\Gamma(s)}}=\ln x\).>j,a
\(\because\quad\;\ln\Gamma\,\)是凸函数,\(\,\Gamma'/\Gamma=(\ln\Gamma)'\) 严格增\((x>0)\),故据\((^*),\) m5
\(\qquad{\large\frac{\Gamma{\hspace{1px}}'(x)}{\Gamma(x)}}< \ln x< {\large\frac{\Gamma{\hspace{1px}}'(x+1)}{\Gamma(x+1)}}\implies\ln(x{\small-1})<{\large\frac{\Gamma{\hspace{1px}}'(x)}{\Gamma(x)}}<\ln x.\)2[y%;
\(\therefore\quad\; 0>{\large\frac{\Gamma{\hspace{1px}}'(x)}{\Gamma(x)}}-\ln x>\ln(x-1)-\ln x=\ln(1-\frac{1}{\large x})\overset{x\to\infty}{\longrightarrow}0.\)2
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2020/12/13 08:30am 
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