Let $k$ be an integer greater than 1. Suppose $a_0 > 0$ and define $a_{n+1} = a_n + \frac{1}{\sqrt[k]{a_n}}$ for $n > 0$. Evaluate $\lim\limits_{n\to\infty} \dfrac{a_n^{k+1}}{n^k}.$
Would the Cesaro-Stolz theorem be useful for this problem? It states that if $(x_n)$ and $(y_n)$ are two sequences of real numbers with $(y_n)$ being strictly positive, increasing, and unbounded, then if $\lim\limits_{n\to\infty} \dfrac{x_{n+1} – x_n}{y_{n+1}-y_n} = L$, then the limit $\lim\limits_{n\to\infty} \dfrac{x_n}{y_n}$ exists and equals L. Clearly $(a_n)$ is a strictly increasing sequence, as can be shown by induction. So maybe I can define $(y_n)$ in terms of $(a_n)$?
Best Answer
Let $b_n=1/a_n$.Obviously $\lim_{n\to\infty}b_n=0,\lim_{n\to \infty }\frac{b_{n+1}}{b_n}=1$,then
\begin{aligned} \lim_{n\to \infty}\frac{a_n^{1+1/k}}{n}&=\lim_{n\to \infty}\frac{(1/b_n)^{1+1/k}}{n} \\& =\lim_{n\to \infty}\left(\frac{1}{b_{n+1}^{1+1/k}}-\frac{1}{b_{n}^{1+1/k}}\right) &\text{(Cesaro-Stolz)} \\&=\lim_{n\to \infty}\left(\frac{b_{n}^{1+1/k}-b_{n+1}^{1+1/k}}{b_{n+1}^{1+1/k}b_{n}^{1+1/k}}\right) \\&=\lim_{n\to \infty}\left(\frac{b_{n}^{1+1/k}(1-(1+b_{n}^{1+1/k})^{-1-1/k})}{b_{n+1}^{1+1/k}b_{n}^{1+1/k}}\right) \\&=\lim_{n\to \infty}\left(\frac{b_{n}^{1+1/k}\left[(1+1/k)b_{n}^{1+1/k}+o(b_{n}^{1+1/k})\right]}{b_{n+1}^{1+1/k}b_{n}^{1+1/k}}\right) \\&=1+\frac{1}{k}. \end{aligned}
Thus $$ \lim_{n\to \infty}\frac{a_n^{k+1}}{n^k}=\left(1+\frac{1}{k}\right)^k $$