Question: Set $x\_1=a$, where $a > 0$ and let $x\_{n+1}= x\_n + \frac{1}{x\_n}$. Determine if the sequence $ \lbrace X\_n \rbrace $ converges or diverges.

I think this sequence diverges since Let $x\_1= a > 0$ and $x\_{n+1}= x\_n + (1/x\_n)$ be given. Let $x\_1=2$ since $a >0$. Then... 

$x\_2 = 2 + \frac{1}{2} = 2.5$

$x\_3 = 2.5 + \frac{1}{2.5} = 2.9$

$x\_4 = 2.9 + \frac{1}{2.9} = 3.2448$

$x\_5 = 3.2448 + \frac{1}{3.2448} = 3.5530$

$x\_6 = 3.5530 + \frac{1}{3.5530} = 3.8344$

$x\_7 = 3.8344 + \frac{1}{3.8344} = 4.0952$

$x\_8 = 4.0952 + \frac{1}{4.0952} = 4.3394$

$x\_9 = 4.3394 + \frac{1}{4.3394} = 4.5698$

$x\_{10} = 4.5698 + \frac{1}{4.5698} = 4.7887$

but others have said it converges so I'm confused on whether it converges or diverges? Can someone please explain.

Answer: 
It is easy to see (by induction) that $x\_{n+1}>x\_n \ge a >0$ for all $n$.

We assume that $(x\_n)$ is convergent and that $l$ is the limmit. Then $l \ge a>0$.

From $x\_{n+1}= x\_n + (1/x\_n)$ we get $l= l+ 1/l$, which is impossible. Hence the sequence is divergent.
As others have shown, $x\_{n+1} \geq x\_n > 0$ for all $n \in \mathbb{N}$. The important point that we will be using is the fact that each term is strictly positive (i.e., non-zero).

Suppose that $(x\_n)$ is convergent. Let $L$ be its limit. $L$ is either zero or non-zero. We shall first show that $L \neq 0$. Consider the sequence $(x\_{n+1}^2)$. If $L = 0$, notice that $\lim\_{n\to\infty}x\_{n+1}^2 = L^2 = 0$. For all $n$,

$$x\_{n+1}^2 = \left(x\_n + \frac{1}{x\_n}\right)^2 = x\_n^2 + \frac{1}{x\_n^2} + 2 \geq 2$$

implying that $\lim\_{n\to\infty}x\_{n+1}^2 \geq 2$, which is a contradiction. Hence, $L \neq 0$.

Since $L \neq 0$,

$$
\lim\_{n\to\infty}x\_n = \lim\_{n\to\infty}x\_{n+1} = \lim\_{n\to\infty}\left(x\_n + \frac{1}{x\_n}\right) = \lim\_{n\to\infty}x\_n + \frac{1}{\lim\_{n\to\infty} x\_n}
$$

The last equality is true because $L \neq 0$ and $x\_n \neq 0$. Hence,

$$
L = L + \frac{1}{L} \Leftrightarrow \frac{1}{L} = 0
$$

which is also a contradiction. Hence $(x\_n)$ is divergent.