Study convergent sequence Study convergent sequence of real numbers given by  
$x_{n+1} = ax_{n} + \frac{b}{x_{n}}, (\forall) n\in N$, where $a, b, x_{0}\in ( 0, \infty)$.
What happens if $a, x_{0}\in (0, \infty)$ but $b<0$?
 A: I am not completely sure about my proof, so please pay attention to the details and tell me if something is wrong. What I will try to prove is that the succession defined by $$x_{n+1}=x_{n}+ \frac{1}{x_{n}}$$ diverges for $x_{0}\leq 1$. In fact we can use induction on n to prove that $$x_{n}>\sum_{1}^n \frac{1}{k},\forall n\geq 1.$$ It is true that $x_{1}=x_{0}+ \frac{1}{x_{0}}>1, (\frac{1}{x_{0}}\geq 1)$. Let us suppose that  $x_{n}>\sum_{1}^n \frac{1}{k},$, then: $$x_{n+1}=x_{n}+ \frac{1}{x_{n}}>\sum_{1}^n \frac{1}{k}+\frac{1}{x_{n}}.$$By induction we now prove that $\frac{1}{x_{n}}\geq \frac{1}{n+1}:$ we know that $x_{0}\leq1$, and suppose $x_{n}\leq n+1$. Now we know that $x_{1} \geq 1$, which implies (the succession is monotonically increasing) $x_{n} \geq 1 \Leftrightarrow \frac{1}{x_{n}} \leq 1$ then: $$x_{n+1}=x_{n}+ \frac{1}{x_{n}} \leq n+2.$$ This proves the assert. Now let's complete the proof for every $a, b, x_{0} >0$. This should not be difficult: first of all let us consider the same succession as before with $x_{0}>1$, can you see that it is the same as choosing it's inverse (say $x'_0=\frac{1}{x_0}$), for which we know the succession diverges? Now let us consider that: $$y_{n+1}=ay_n + \frac{b}{y_n} \geq cx_{n+1}, c=\min \lbrace a, b\rbrace.$$This should prove that your succession diverges.
