Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? 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.
 A: 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.
A: 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.
A: Assume the sequence converges i.e. $\displaystyle\lim_{n\to \infty} x_n = C< \infty.$
Then, as $n\to\infty$  we have $\,x_{n}\approx x_{n+1}$.
Taking limit of $x_{n+1}=x_n+\frac{1}{x_n}$ we get
$$
\lim_{n\to\infty} x_{n}= \lim_{n\to\infty} \left(x_n+\frac{1}{x_n} \right)\implies C =  C + \frac{1}{C} \iff \frac{1}{C} = 0 \quad  \text{ – } \text{ contradiction}
$$
The last expression results in contradiction even if $\,C\sim 0$.
A: HINT: Show that $x_{n + \lceil 2 x_n \rceil} > 2 x_n$.
A: We can see the series $x_{n+1}$ is monotone increasing as $n\rightarrow \infty$ using the Monotone Convergence Theorem:
Monotone Increasing $\Rightarrow x_{n+1}>x_{n}, \forall n$
$\Rightarrow x_n+\frac{1}{x_n}>x_n$
$\Rightarrow \frac{1}{x_n} > 0$
Which is true since $x_1=a$ and $a>0$.
So now we know we have an increasing sequence. Lets assume some limit $x$ exists, which implies $x_{n+1}=x_n$, so we solve to 'find' our limit:
$x=x+\frac{1}{x}\Rightarrow x^2=x^2+1$, leading to a contradiction. 
Therefore the limit does not exist and the sequence diverges.   
A: 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.
