Showing a recursive sequence isn't bounded $a_{n+1}=a_n+\frac 1 {a_n}$

Notice $$a_{n+1} = a_n + \frac{1}{a_n} \quad\implies\quad a_{n+1}^2 = a_n^2 + 2 + \frac{1}{a_n^2} \ge a_n^2 + 2$$ This implies for any $n$, $$a_n^2 \ge a_1^2 + 2(n-1) = 2n-1$$ As a result, $a_n$ diverges at least as fast as $\sqrt{2n-1}$ as $n\to\infty$.

Hint: The sequence is obviously increasing. So if it is bounded, then it has a limit $b\ge 1$. Thus $b=\lim_{n\to\infty} a_{n+1}=\cdots$.

Somewhat of a funny proof.

Suppose for contradiction that $a_n$ is bounded by some positive $M$

Then the series $\displaystyle \sum (a_{n+1}-a_n)=\sum\frac{1}{a_n}$ has bounded partial sums (and positive general term).

Thus, the series $\displaystyle \sum\frac{1}{a_n}$ converges, which implies $a_n\to \infty$

Contradiction.