Let $x_1=a>0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$. Check whether the following sequence converges or diverges. 
Let $x_1=a>0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$.
  Check whether the following sequence converges or diverges.

When I was in UG my teacher used derivative test for monotonicity. 
$f(x)=x+\frac{1}{x}, f'(x)=1-\frac{1}{x^2}>0(x>1).$ So, $f(x)$ is increasing. How to prove the sequence is monotonic? Differentiation is coming after the sequences and series. By AM-GM inequality sequence is bounded below. $x_{n+1}=x_n+\frac{1}{x_n}\ge 2\sqrt{x_n.\frac{1}{x_n}}=2  \forall n\in \mathbb N$. How can I judge whether the sequence bounded above or not? Please help me.
 A: Rearranging we have $x_{n+1}-x_n=\frac{1}{x_n}$. Multiplying by $x_{n+1}+x_n$ we get 
$$x^2_{n+1}-x^2_n=1+\frac{x_{n+1}}{x_n} \geq 1$$
Telescoping the sum we obtain that,
$$x^2_{n+1}-x_1^2\geq n$$
Hence $x_n$ is unbounded.
A: The sequence is increasing indeed. 
Since $$x_{n+1} - x_n = \frac{1}{x_n} > 0$$
We can show that it is not bounded. Otherwise, it converges, say to $l$.
Then we have $$ l = l + 1/l,$$ which is a contradiction.
A: Extending Clark's answer.
$x_{n+1}^2
=x_n^2+2+\dfrac1{x_n^2}
$
so
$x_{n+1}^2-x_n^2 \ge 2$.
Summing,
$x_{n+1}^2-x_1^2 \ge 2n$
so
$x_{n+1}^2
\ge x_1^2 + 2n
= 2n+a^2
$
so
$x_{n+1} 
\ge \sqrt{2n+a^2}
\gt \sqrt{2n}
$.
Therefore
$x_{n+1}
=x_n+\frac{1}{x_n}
\le x_n+\frac{1}{\sqrt{2(n-1)}}
$
or
$x_{n+1}- x_n
\le \frac{1}{\sqrt{2(n-1+a^2)}}
\lt \frac{1}{\sqrt{2(n-1)}}
$.
Summing,
$\begin{array}\\
x_{m+1}- x_2
&=\sum_{n=2}^m (x_{n+1}- x_n)\\
&\lt \sum_{n=2}^m\frac{1}{\sqrt{2(n-1)}}\\
&= \dfrac1{\sqrt{2}}\sum_{n=1}^{m-1}\frac{1}{\sqrt{n}}\\
&= \dfrac1{\sqrt{2}}(1+\sum_{n=2}^{m-1}\frac{1}{\sqrt{n}})\\
&< \dfrac1{\sqrt{2}}(1+\int_1^m \dfrac{dt}{t^{1/2}})\\
&= \dfrac1{\sqrt{2}}(1+\dfrac{t^{1/2}}{1/2}|_1^m)\\
&= \dfrac1{\sqrt{2}}(1+2(\sqrt{m}-1))\\
&= \dfrac1{\sqrt{2}}(2\sqrt{m}-1)\\
&= \sqrt{2m}-\dfrac1{\sqrt{2}}\\
\end{array}
$
so that
$\sqrt{2m}
\lt x_{m+1}
\lt \sqrt{2m}+x_2-\dfrac1{\sqrt{2}}
$.
Extending this,
I am sure that we can prove that
$\lim_{m \to \infty} (x_m-\sqrt{2m})
$
exists
(and I have a feeling that
I already have),
but I'll leave it at this.
A: Hint: Assume for the sake of contradiction that the sequence is bounded above by some threshold $\lambda$. What can we say about the differences $x_{n+1}-x_n$? On one hand, $x_{n+1}-x_n = 1/x_n \geq 1/\lambda$, but...
