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.
We have $$x\_{N+1}-x\_1=\sum\_{n=1}^N x\_{n+1}-x\_n =\sum\_{n=1}^N \frac1{x\_n}$$ so $x\_n$ converges if and only if the rightmost series converges. 

But the series converges only if $x\_n$ diverges to $\pm\infty $, impossible: both the sequence and the series must diverge.