recursive sequence $x_{n+1}=\frac{1}{x_n}$ $x_{n+1}=\frac{1}{x_n}$, $x_1>1$
Find if the function converges and diverges and then prove it.
If we try and find the limit we get 1 or -1. 1 does not work because then $x_1$ and $x_n$ contradict each other. -1 works for showing its the lowest bound. I'm not sure about this proof for proving it's monotonically decreasing:
$x_n>-1$
which implies ${x_n^2}>1$ whcih implies $x_{n}>\frac{1}{x_n}$ and so $x_n>x_{n+1}$
therefore decreasing sequence
 A: Suppose $x_1=a$. Then $x_n = a$ if $n$ is odd and $x_n=\frac{1}{a}$ if $n$ is even.
Define $\varepsilon = \frac{a-1}{a}$ (which is positive by the assumption that $a>1$. It follows that for any $n \in \mathbb{N}$ we have $|x_n-x_{n+1}| = a-\frac{1}{a} = \frac{a-1}{a} = \varepsilon \ge \varepsilon$. Thus the negation of the definition of convergence is satisfied, so the sequence diverges.
A: Notice that the sequence will follow
$$ x_n = \begin{cases} x_1 & \text{if $n$ is odd} \\[2ex] \frac{1}{x_1} & \text{if $n$ is even} \end{cases} $$
Hence, the sequence can only converge if $x_1=\frac{1}{x_1} \Leftrightarrow x_1 = \pm 1$. Otherwise, and in particular if $x_1>1$ it does not converge, instead it oscillates perpetually between $x_1$ and $\frac{1}{x_1}$. 
A: As pointed out in the comments, let's look at an example: 

Let $x_1 = 2$. Then, $x_2 = \frac{1}{x_1} = \frac{1}{2}$, and $x_3 = \frac{1}{x_2} = \frac{1}{\frac{1}{2}} = 2$. Let's compute two more: $x_4 = \frac{1}{x_3} = \frac{1}{2}$ and, finally, $x_ 5 = \frac{1}{x_4} = \frac{1}{\frac{1}{2}} = 2$. 

Note that if a sequence of real numbers $(a_n)_n$ converges to $a$, then every subsequence must converge to $a$. However, as you will see, we have a subsequence $(a_{n_k})_k$ namely the one given by $2, 2, 2, ...$ which converges to $2$ and another subsequence  $(a_{n_l})_l$ given by $1/2, 1/2, ...$ which converges to $1/2$. since $1/2 \neq 2$ the sequence $(a_n)_n$ does not converge.  
See the other answer, which is a generalization. 
A: From comments it appears that you think to diverge means something other to fail to converge.  Maybe that is the whole problem here.
Look at a concrete case:
\begin{align}
x_1 & = 5 \\[5pt]
x_2 & = 1/5 \\[5pt]
x_3 & = 5 & \text{At this point the pattern should become clear.} \\[5pt]
x_4 & = 1/5 \\[5pt]
x_5 & = 5 \\[5pt]
x_6 & = 1/5 \\[5pt]
& \vdots
\end{align}
This does not converge.  I.e. it diverges.
If you start with $x_1=1$ or $x_1=-1$, then it converges.
