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发表于 2020-10-17 06:19
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\(0< x< 1\) 时\(\,0< {\large\frac{\ln(1+x)}{x}}\small=1-(\dfrac{x}{2}-\dfrac{x^2}{3})-\cdots < 1\)
\(x\ge 1\) 时\(\,0<{\large\frac{\ln(1+x)}{x}}=\ln(1+x)^{\frac{1}{x}}< \ln e=1\) 所以
\(\,{\small\dfrac{a_{n+1}}{a_n}}< 1,\; a_{n+1}< a_n,\;\{a_n\}\,\)递减有下界. 极限\(\,A\ge 0\).
\( \therefore\,\displaystyle\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\ln(1+a_n)\implies A=\ln(1+A)\)
\(\because\;A>0\implies A>\ln(1+A).\;\;\therefore\; A=0. \;\displaystyle\lim_{n\to\infty}a_n = 0\)
\(\displaystyle\lim_{n\to\infty}na_n=\lim_{n\to\infty}{\small\dfrac{n}{a_n^{-1}}}\overset{stolz}{=}\lim_{n\to\infty}{\small\frac{1}{a_{n+1}^{-1}-a_n^{-1}}}=\lim_{n\to\infty}{\small\frac{a_na_{n+1}}{a_n-a_{n+1}}}\)
\(=\displaystyle\lim_{n\to\infty}{\small\frac{a_na_{n+1}}{a_n-\ln(1+a_n)}}=\lim_{n\to\infty}{\small\frac{a_na_{n+1}}{a_n-(a_n-\frac{1}{2}a_n^{2}+O(a_n^3))}}\)
\(\displaystyle=\lim_{n\to\infty}{\small\frac{2a_{n+1}}{a_n}}=\lim_{n\to\infty}{\small\frac{2(a_n+O(a_n^2))}{a_n}}=2\)
\(\displaystyle\lim_{n\to\infty}{\small\frac{n(na_n-2)}{\ln n}}=\lim_{n\to\infty}{\small\frac{na_n(n-\frac{2}{a_n})}{\ln n}}=2\lim_{n\to\infty}{\small\frac{n-2/a_n}{\ln n}}\)
\(\displaystyle\overset{stolz}{=}2\lim_{n\to\infty}{\small\frac{1-2/a_{n+1}+2/a_n}{\frac{a_n}{na_n}\ln(1+\frac{1}{n})^n}}=4\lim_{n\to\infty}{\small\frac{(a_n+2)a_{n+1}-2a_n}{a_n^2a_{n+1}}}\)
\(\displaystyle=4\lim_{n\to\infty}{\small\frac{(a_n+2)a_{n+1}-2a_n}{a_n^3}}=4\lim_{n\to\infty}{\small\frac{\frac{1}{6}a_n^3+O(a_n^4)}{a_n^3}}=\small\frac{2}{3}\) |
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发表于 2020-10-22 09:17