Showing a recursive sequence isn't bounded $a_{n+1}=a_n+\frac 1 {a_n}$ 
Show the sequence isn't bounded: $a_1=1$, $a_{n+1}=a_n+\frac 1 {a_n}$.

Proof by contradiction: Let $M>0$ such that $\forall n: |a_n|< M$. 
Let $\epsilon >0 $ and for some $n=N, \epsilon: a_N=M-\epsilon<M $ pluging that in the recursion: $a_{N+1}=M-\epsilon+\frac 1 {M-\epsilon}>M>M-\epsilon$. Contradiction.
I wondered if I could suppose about the boundary that $\forall n: |a_n|\le M$ ? The proof would basically be the same only I could drop the epsilon.
 A: Hint: The sequence is obviously increasing. So if it is bounded, then it has a limit $b\ge 1$. Thus $b=\lim_{n\to\infty} a_{n+1}=\cdots$. 
A: 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.
A: Hint: First, show by induction that $a_n > 0$ for all $n$
Now, suppose $a_n$ is bounded, i.e. $a_n \le M$ for some $M > 0$. 
Then, $\dfrac{1}{a_n} \ge \dfrac{1}{M}$, and so, $a_{n+1} = a_n + \dfrac{1}{a_n} \ge a_n + \dfrac{1}{M}$. 
Now, can you show how this yields a contradiction?
A: In addition to what was said before, you can prove by induction that $\ln n\leq a_n\leq n$. It holds for $n=1,2,3$ and if it holds for $n$, then $$a_{n+1}=a_n+\frac{1}{a_n}\geq a_n+\frac{1}{n}\geq \ln n + \frac{1}{n}\geq \int_1^n \frac{1}{x} \,dx+\int_n^{n+1} \frac{1}{x} dx=\ln (n+1)$$
and similarly, you show that $a_{n}+\frac{1}{a_n}\leq n+\frac{1}{\ln n}\leq n+1$.
A: 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 $n\to\infty$.
