How fast does a_{n+1} = a_n + 1/a_n grow?

Squaring. $a_{n+1}^2=a_n^2+2+\dfrac1{a_n^2}$. All $a_n\ge1>0$.

Lower bound. Since $\frac1{a_n^2}>0$, $a_{n+1}^2>a_n^2+2$, so $a_n^2\ge a_1^2+2(n-1)=2n-1$.

Upper bound. Summing the squared recurrence from $1$ to $n-1$,
$$a_n^2=1+2(n-1)+\sum_{k=1}^{n-1}\frac1{a_k^2}\le2n-1+\sum_{k=1}^{n-1}\frac1{2k-1},$$
using $a_k^2\ge2k-1$. Since $\frac1{2k-1}\le\frac1k$ for $k\ge1$, the last sum is at most $H_{n-1}\le1+\ln(n-1)$ for $n\ge2$. Thus
$$a_n^2\le2n-1+1+\ln n=2n+\ln n.$$

2. Dividing by $2n$: $1-\frac1{2n}\le\frac{a_n^2}{2n}\le1+\frac{\ln n}{2n}$, and both bounds tend to $1$. Hence $\frac{a_n}{\sqrt{2n}}\to1$.

3. $a_{100}^2\ge199>196=14^2$ and $a_{100}^2\le200+\ln100<205<225=15^2$. So $14<a_{100}<15$.

Lesson. For a recurrence $a_{n+1}=a_n+\varepsilon_n$ with small increments, look at $a_n^2$ so that the main term becomes linear.