3.3.14 ** The sequence $\left\{a_{n}\right\}$ is defined by the following conditions: $a_{1}=1, a_{n+1}=a_{n}+\frac{1}{a_{n}}, n \geqslant 1$, find the integer part of $a_{100}$.
From the given, we know that $a_{n+1}^{2}=a_{n}^{2}+2+\frac{1}{a_{n}^{2}}$, which means $a_{n+1}^{2}-a_{n}^{2}=2+\frac{1}{a_{n}^{2}}$. Therefore,
$$
a_{100}^{2}-a_{99}^{2}=2+\frac{1}{a_{99}^{2}}, a_{99}^{2}-a_{98}^{2}=2+\frac{1}{a_{98}^{2}}, \cdots, a_{2}^{2}-a_{1}^{2}=2+\frac{1}{a_{1}^{2}} .
$$

By adding up all these equations, we get
$$
a_{100}^{2}-a_{1}^{2}+2 \times 99+\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+\cdots+\frac{1}{a_{99}^{2}}=200+\frac{1}{a_{2}^{2}}+\frac{1}{a_{3}^{2}}+\cdots+\frac{1}{a_{99}^{2}} .
$$

Since $a_{n+1}-a_{n}=\frac{1}{a_{n}}>0$, the sequence $\left\{a_{n}\right\}$ is an increasing sequence, and when $n \geqslant 2$, $a_{n} \geqslant a_{2}=$
2. Thus, $200<a_{100}^{2} \leqslant 200+\frac{1}{2^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{2}}$, which means $200<a_{100}^{2}<225,14<a_{100}<$
15, hence $\left\lfloor a_{100}\right\rfloor=14$.