3. The sequence $a_{1}, a_{2}, \ldots$ is defined by the equalities

$$
a_{1}=100, \quad a_{n+1}=a_{n}+\frac{1}{a_{n}}, \quad n \in \mathbb{N}
$$

Find the integer closest to $a_{2013}$.
# Answer: 118.

Solution.

\[
\begin{aligned}
a_{2013}^{2}=\left(a_{2012}+\frac{1}{a_{2012}}\right)^{2}=a_{2012}^{2}+2+\frac{1}{a_{2012}^{2}}=a_{2011}^{2} & +2 \cdot 2+\frac{1}{a_{2011}^{2}}+\frac{1}{a_{2012}^{2}}=\ldots \\
& =a_{1}^{2}+2 \cdot 2012+\frac{1}{a_{1}^{2}}+\ldots+\frac{1}{a_{2011}^{2}}+\frac{1}{a_{2012}^{2}}
\end{aligned}
\]

Therefore, on one hand,

\[
a_{2013}^{2}>a_{1}^{2}+2 \cdot 2012=10000+4024=14024>118^{2}=13924
\]

on the other hand,

\[
a_{2013}^{2}<a_{1}^{2}+2 \cdot 2012+\frac{2012}{100^{2}}<14024+1<118.5^{2}
\]

Therefore, \(118<a_{2013}<118.5\), and the nearest integer is 118.