(The eighth question of the 2017 Peking University Youte Test)
The sequence \(\{a_n\}\) satisfies \(a_1=1\), \(a_{n+1}=a_n+\dfrac{1}{a_n}\), if \(a_{2017}\in ( k,k+1)\), where \(k\in\mathbb N^{\ast} \), then the value of \( k\) is ______
A. \(63\)
B. \(64\)
C. \(65\)
D. \(66\)
Answer: A
Hint : For the above recursion we actually have $\lim\limits_{n\to+\infty}\dfrac{a_n}{\sqrt n}=\sqrt 2.$
If the sequence recursion is changed to $ a_{n+1}=a_n+\dfrac{1}{a^2_n}$
We have the following estimates: $\lim\limits_{n\to+\infty}\dfrac{a_n}{\sqrt[3]{n} }=\sqrt[3]{3}$