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楼主 |
发表于 2020-10-5 17:07
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题：试证\(\underset{\,}{\;}\tau_n\sim\frac{1}{3}\ln n\;\;({\small\tau_n=n-}\frac{2}{a_n},{\small\,a_{n+1}=\ln(1+a_n),\,a_1>0})\)
证：\(\;\because\;0< \ln(1+x) < x\;\;(x>0),\)
\(\qquad\therefore\;\;0< a_{n+1}=\ln(1+a_n)< a_n\;\;(a_1>0)\)
\(\qquad\therefore\;\;a_n\hspace{-2px}{\small\searrow}\; a\ge 0,\;\;a=\ln(1+a),\;{\displaystyle\lim_{n\to\infty}}a_n = a=0\)
\(\qquad\displaystyle{\small\lim_{n\to\infty}}na_n{\small=\lim_{n\to\infty}}{\scriptsize\frac{\Delta n}{\Delta a_n^{-1}}}\small=\lim_{n\to\infty}\frac{a_na_{n+1}}{a_n-a_{n+1}}=\lim_{n\to\infty}\frac{a_n^2}{\frac{1}{2}a_n^2+O(a_n^3)}=2\)
\(\therefore\quad\tau_{n+1}-\tau_n=1-2\big(\frac{1}{\ln(1+a_n)}-\frac{1}{a_n}\big)=\frac{1}{6}a_n+O(a_n^2)\sim\frac{1}{3n}\underset{\,}{,}\)
\(\qquad\tau_n\sim\frac{1}{3}H_n\sim\frac{1}{3}\ln n,\quad{\large\frac{n(na_n-2)}{\ln n}}={\large\frac{na_n\tau_n}{\ln n}}\sim{\large\frac{na_n}{3}}\sim{\large\frac{2}{3}}.\) |
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发表于 2020-10-5 10:05
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