Let $$x_n=\sqrt{2}+\sqrt[3]{\dfrac{3}{2}}+\ldots+\sqrt[n+1]{\dfrac{n+1}{n}}, \qquad n=1,2,3,\ldots$$
Evaluate $$\lim_{n \to \infty} \dfrac{x_n}{n}.$$
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam
Solution:
Let $y_n=n$. Since $(y_n)_{n \geq 1}$ is strictly monotone and divergent sequence and
$$\lim_{n \to \infty} \dfrac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n \to \infty} \left(1+\dfrac{1}{n+1}\right)^{\frac{1}{n+2}}=1,$$ then by the Stolz-Cesaro Theorem we have $$\lim_{n \to \infty} \dfrac{x_n}{n}=1.$$
$$\lim_{n \to \infty} \dfrac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n \to \infty} \left(1+\dfrac{1}{n+1}\right)^{\frac{1}{n+2}}=1,$$ then by the Stolz-Cesaro Theorem we have $$\lim_{n \to \infty} \dfrac{x_n}{n}=1.$$
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