问题：$x_{1},x_{2}>0$,$\{x_{n}\}$满足
\[ x_{n+1}=x_{1}^{\frac{1}{n}}+x_{2}^{\frac{1}{n}}+\cdots+x_{n}^{\frac{1}{n}}\]
求
\[ \lim_{n\to\infty}\frac{x_{n}-n}{\ln{n}}\]
解：设$a=\max\{x_{1},x_{2},4\}$,由Holder不等式。可以得到
\[ x_{n+1}^{n}\leq \left(\sum_{k=1}^{n}x_{k}\right)\cdot n^{n-1}\]
\[ x_{n+1}\leq  \left(\sum_{k=1}^{n}x_{k}\right)^{\frac{1}{n}}\cdot n^{\frac{n-1}{n}}\]
这样，用数学归纳法可以证明
\[ 0<a_{n}<an \]
而Taylor公式告诉我们
\[ x_{k}^{\frac{1}{n}}=1+\frac{1}{n}\ln x_{k}+o\left(\frac{1}{n}\right)\]
于是
\[ x_{n}=n-1+\frac{1}{n-1}\ln \prod_{k=1}^{n-1}x_{k}+o\left(\frac{1}{n-1}\right)\]
\begin{align*}
\lim_{n\to\infty}\frac{x_{n}-n}{\ln{n}}&=\lim_{n\to\infty}\frac{\frac{1}{n-1}\ln \prod_{k=1}^{n-1}x_{k}-1 }{\ln{n}}\\
&=\lim_{n\to\infty}\frac{\ln \prod_{k=1}^{n-1}x_{k}-(n-1)}{(n-1)\ln{n}}\\
&=\lim_{n\to\infty}\frac{\ln x_{n}-1}{n\ln(n+1)-(n-1)\ln{n}}\qquad (O.Stolz)\\
&=\lim_{n\to\infty}\frac{\ln x_{n}-1}{\ln(n+1)}\\
&=\lim_{n\to\infty}\frac{\ln \frac{x_{n+1}}{x_{n}}}{\frac{1}{n+1}}\qquad (O.Stolz)\\
&=1
\end{align*}
最后一步只要注意到
\[ \frac{x_{n+1}}{x_{n}}=\frac{n+\frac{1}{n}\ln\prod_{k=1}^{n}x_{k}+o\left(\frac{1}{n}\right)}{n-1+\frac{1}{n-1}\ln \prod_{k=1}^{n-1}x_{k}+o\left(\frac{1}{n-1}\right)}\]
利用上面的放缩，可以看到
\[ \ln\left(\frac{x_{n+1}}{x_{n}}\right)\sim \ln\left(1+\frac{1}{n-1}\right)\quad (n\to+\infty) \]