Recursive sequence does not converge Define $x_n$ recursively as follows: $x_1=1$, $x_{n+1}=x_n+\frac{1}{x_n}$. We are asked to show this sequence is not convergent. Here's my attempt.
Since $x_1=1>0$, and for each $n \in \mathbb{N}, |x_n|>0$, we must have $|\frac{1}{x_n}|>0$ and hence $|x_n|<|x_{n+1}|$. Which means $\{x_n\}$ is a monotone sequence and to show it doesn't converege, it is sufficient to show that it isn't bounded. Let us assume $\lbrace x_n \rbrace$ is bounded and let $M=\sup \lbrace x_n \rbrace$. 
By supremum property of $\mathbb{R}$, given $\epsilon> \frac{1}{M} >0, \exists k \in \mathbb{N},  x_k>M-\epsilon$. Say $x_k=\delta>M-\frac{1}{M}$. Note that $M \geq \delta \implies  \frac{1}{\delta} \geq \frac{1}{M}$. Also, $M-\frac{1}{M}< \delta \leq M$ $\implies$ $M < \delta +\frac{1}{M} \leq M+\frac{1}{M}$, and from this we get $M< \delta+\frac{1}{M}< \delta+\frac{1}{\delta}$. But, $x_{k+1}=\delta+\frac{1}{\delta}$, which is a contradiction, hence the sequence is unbounded. 
Does this look okay?
Edit: Changed the definition of M as per the comment.
 A: The idea is fine, but some of the details are wrong. First, by your definition $\sup\{y\in\Bbb R:y\ge x_n\}=+\infty$, which definitely isn’t what you want; from what follows it appears that you want to define $M$ to be the assumed least upper bound of the sequence, i.e., $M=\sup\{x_n:n\in\Bbb Z^+\}$.
Your $\epsilon$ is never actually used. Moreover, if $\epsilon>\frac1M$, choosing $k\in\Bbb Z^+$ so that $x_k>M-\epsilon$ does not ensure that $x_k>M-\frac1M$, since $M-\frac1M>M-\epsilon$. Just observe that the definition of $M$ ensures that there is a $k\in\Bbb Z^+$ such that $x_k>M-\frac1M$. There is no need to rename it $\delta$: that just introduces an unnecessary symbol. Then you can complete the argument roughly as you did, but a bit more efficiently: $x_k<M$, so $\frac1{x_k}>\frac1M$, and therefore
$$x_{k+1}=x_k+\frac1{x_k}>\left(M-\frac1M\right)+\frac1M=M\;,$$
contradicting the definition of $M$.
A: Assuming that the sequence converges, it converges to some $x_\infty$ such that
$$x_\infty=x_\infty+\frac1{x_\infty}.$$
This proves that convergence is impossible.
