Let a sequence be defined by $a_1=1$ and $a_{n+1}=a_n+\frac{1}{a_n}$ for $n\ge 1$. Find the integer part of $a_{100}$.

Solution

1. Given $a_1=1$ and $a_{n+1}=a_n+\frac{1}{a_n}$, squaring both sides yields $a_{n+1}^2=a_n^2+2+\frac{1}{a_n^2}$.
2. Summing this relation from $n=1$ to $99$ gives $$\begin{array}{rl}{a}_{100}^2& =a_1^2+198+\sum _{k=1}^{99}\frac{1}{a_k^2}\\ & =199+\sum _{k=1}^{99}\frac{1}{a_k^2}\end{array}$$.
3. Since $a_n$ is increasing, $a_n^2\ge 1$, so $199<a_{100}^2<199+{\sum }_{k=1}^{99}\frac{1}{k}$, which implies $14<a_{100}<15$.
4. Hence the answer is (B).