设 $\sed{a_n}$, $\sed{b_n}$ 都是正数列, 满足 $$\bex \lim_{n\to\infty}\frac{b_n}{n}=0\mbox{ 及 } \lim_{n\to\infty}b_n\sex{\frac{a_n}{a_{n+1}-1}-1}=\lambda>0. \eex$$ 求证:
(1) $\dps{\lim_{n\to\infty}a_n=0}$;
(2) 级数 $\dps{\sum_{n=1}^\infty a_n}$ 收敛.
证明: 由 $$\bex \lim_{n\to\infty} n\sex{\frac{a_n}{a_{n+1}}-1} =\lim_{n\to\infty}\sez{b_n\sex{\frac{a_n}{a_{n+1}}-1}}\cdot\frac{1}{b_n/n} =+\infty \eex$$ 知当 $n$ 充分大时, $$\bex n\sex{\frac{a_n}{a_{n+1}}-1}\geq {\color{red}3}. \eex$$ 而 $$\bex \frac{a_{n+1}}{a_n} &\leq&\frac{n}{n+3}\\ &=&\frac{1}{1+\frac{2}{n}+\frac{1}{n}}\\ &\leq&\frac{1}{1+\frac{2}{n}+\frac{1}{n^2}}\\ &=&\frac{1}{\sex{1+\frac{1}{n}}^2}\\ &=&\frac{n^2}{(n+1)^2}. \eex$$ 于是 $$\bex a_{n+1}&\leq&\frac{n^2}{(n+1)^2}a_n\\ &\leq&\frac{n^2}{(n+1)^2}\cdot\frac{(n-1)^2}{n^2}a_{n-1}\\ &\leq&\cdots\\ &\leq&\frac{1}{(n+1)^2}a_1. \eex$$ 这说明 $\dps{\sum_{n=1}^\infty a_n}$ 收敛. 特别的, $\dps{\lim_{n\to\infty}a_n=0}$.