设$k$是一个大于$1$的整数，且$x_0>0$,满足$x_{n+1}=x_{n}+\frac{1}{\sqrt[k]{x_{n}}}$,求
\[ \lim_{x\to\infty}\frac{x_{n}^{k+1}}{n^{k}}\]
求证
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{k+1}}{n^{k}}-\left(\frac{k+1}{k}\right)^{k}\right)=\frac{1}{2}\left( \frac{k+1}{k}\right)^{k-1}\]
证明：
我们将要证明
\[  \lim_{n\to\infty}\frac{x^{k+1}_{n}}{n^{k}}=\left(\frac{k+1}{k}\right)^{k}\]
为此，只要证明
\[  \lim_{n\to\infty}\frac{x^{1+\frac{1}{k}}_{n}}{n}=\frac{k+1}{k}\]
而归纳得到
\[ x_{n+1}>x_{n},x_{n}>0\]
因此，$x_{n}$的单调递增序列。设$A=\lim_{n\to+\infty}x_{n}$,则有$A=+\infty$.说明$x_{n}\to+\infty$,$n\to+\infty$.
\[ (x_{n+1}^{1+\frac{1}{k}}-x_{n}^{1+\frac{1}{k}})=x_{n}^{1+\frac{1}{k}}\left(\left(\frac{x_{n+1}}{x_{n}} \right)^{1+\frac{1}{k}}-1\right)=x_{n}^{1+\frac{1}{k}}\left(\exp\left[\left(1+\frac{1}{k}\right)\ln\left(1+\frac{1}{x_{n}^{1+\frac{1}{k}}} \right) \right]-1 \right)\sim \frac{k+1}{k}(n\to+\infty)\]
所以，由O.Stolz定理
\[  \lim_{n\to\infty}\frac{x^{1+\frac{1}{k}}_{n}}{n}=\frac{k+1}{k}\]
对于加强版本，我们先看一个事实，对任意一个收敛的序列，比如说
\[ \lim_{n\to\infty}a_{n}=A \]
那么，
\[ \lim_{n\to\infty}\frac{(a_{n}^{k}-A^{k})}{a_{n}-A}=kA^{k-1}\]
这样，设$a_{n}=\frac{x_{n}^{\frac{k+1}{k}}}{n}$,就有$\displaystyle \lim_{n\to\infty}a_{n}=\frac{k+1}{k}=A$.
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}^{k}-A^{k})= \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}-A)\cdot k\cdot \left(\frac{k+1}{k}\right)^{k-1}\]
于是，只要证
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}-A)=\frac{1}{2k}\]
就是
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{\frac{k+1}{k}}}{n}-A\right)=\frac{1}{2k}\]
\[ \frac{n}{\ln{n}}\left(\frac{x_{n}^{\frac{k+1}{k}}}{n}-A\right)=\frac{x_{n}^{\frac{k+1}{k}}-nA}{\ln{n}}\]
这时，可以用O.Stolz定理了。就有
\begin{align*}
&\lim_{n\to\infty}\frac{x_{n}^{\frac{k+1}{k}}-nA}{\ln{n}}\\
&\overbrace{=}^{O.Stolz}\lim_{n\to\infty}\frac{x_{n+1}^{1+\frac{1}{k}}-x_{n}^{1+\frac{1}{k}}-A}{\ln\left(1+\frac{1}{n} \right)}\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(\left(\frac{x_{n+1}}{x_{n}}\right)^{1+\frac{1}{k}}-1\right) -A\right]\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(e^{\left(1+\frac{1}{k}\right)\ln\left(1+\frac{1}{x_{n}^{1+\frac{1}{k}}}\right) } -1\right)-A\right]\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(e^{\left(1+\frac{1}{k}\right)\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}+o\left( \left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right) \right) }-1 \right) -A\right]\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(\left(\frac{k+1}{k}\right)\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right)+\frac{1}{2}\left(\frac{k+1}{k}\right)^{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right)^{2} \right)-A\right]\\
&=\lim_{n\to\infty}\frac{k+1}{2k^{2}}\frac{n}{x_{n}^{1+\frac{1}{k}}}\\
&=\frac{1}{2k}
\end{align*}
因此，命题得证。