$\displaystyle let \ x_1=a>0 \ and \ x_{n+1} = x_n+ \frac{1}{x_n}, \ n \in N .$
$\displaystyle Does \ x_n \ converge \ ? $
As $\displaystyle a> 0$ , it's easy to show inductively that $\displaystyle x_n\geq 2\,\,\forall\,2\leq n\in\mathbb{N}$.
If the sequence converged to a finite limit $\displaystyle \alpha$, which must be $\displaystyle \alpha \neq 0$ by the above , then using arithmetic of limits we'd get $\displaystyle \lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n-1}+\frac{1}{\lim_{n\to\infty}x_{n-1}}\Longrightarrow\,\alpha=\alpha+\frac{1}{\alpha} \,\Longrightarrow \alpha=1$ , and b the above it's clear that this can't be so the sequence never converges to a finite limit.
Tonio