Determine whether $(x_n)$ converges or diverges

Question

Given $x_1 := a > 0$ and $x_{n+1} := x_n + \frac{1}{x_n}$ for $n \in \mathbb{N}$, determine whether $(x_n)$ converges or diverges.

Since $x_1 > 0$, it seems obvious that the sequence is strictly increasing and always positive because we are always adding a positive number to each subsequent element in the sequence.

The trickier part is to show whether $(x_n)$ is bounded or not. If $(x_n)$ is bounded, then $\exists M \in \mathbb{R}$ such that

\begin{align}|x_n| \leq M ~\forall n \in \mathbb{N}\tag{1} \end{align}

Alternatively, I think this means that $M$ could be a supremum of the sequence and another way to rewrite $(1)$ is given any $\epsilon > 0$

\begin{align} x_n + \epsilon \leq M \tag{2}\end{align}

Choose $\epsilon = \frac{1}{x_n}$. Then

\begin{align} x_{n+1} = x_n + \frac{1}{x_n} \leq M \implies x_n + \epsilon \leq M \tag{3}\end{align}

Thus I conclude that the inequality holds $\forall n \in \mathbb{N}$ and that $(x_n)$ is convergent. Thus by the Monotone Sequence Convergence Theorem, $(x_n)$ is convergent.

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The part I am not sure about is my reasoning to show that $x_n$ being bounded is correct or not.

Answer 1

As $f(x)=x+\frac{1}{x}$ is continuous, if $\{x_n\}$ would converge to $l$, we would have

$$l=l+\frac{1}{l}$$

As this equality is not possible in $\mathbb R$, the sequence is not convergent.

Answer 2

If $\lim_{n\to\infty}x_n=l$ for some $l\in\mathbb R$ then, since each $x_n$ is greater than or equal to $a$, $l\geqslant a>0$. But then\begin{align}l&=\lim_{n\to\infty}x_n\\&=\lim_{n\to\infty}x_{n+1}\\&=\lim_{n\to\infty}x_n+\frac1{x_n}\\&=l+\frac1l.\end{align}This is impossible, of course. So, the sequence diverges and, since it is monotonic, it is unbounded.