Establishing the convergence or divergence of the sequences $(x_n)$ 
Establish the convergence or the divergence of the sequence $$(x_n) = \frac{(-1)^n n}{n+1}$$

At the moment all I can conclude is that $$(x_n) = \frac{(-1)^n n}{n+1} < (-1)^n\left(\frac{n}{n}\right) = (-1)^n$$
So $(x_n)$ is bounded above by $1$ and below by $-1$. How can I show that $(x_n)$ is convergent or isn't convergent?
 A: You can't, since it isn't. The subsequence $(x_{2n})_{n\in\mathbb N}$ converges to $1$, whereas the subsequence $(x_{2n-1})_{n\in\mathbb N}$ converges to $-1.$
A: Another way to see
that the sequence diverges
is to note that
$\begin{array}\\
x_{2n}-x_{2n+1}
&=\frac{(-1)^{2n} 2n}{2n+1}-\frac{(-1)^{2n+1} 2n+1}{2n+2}\\
&=\frac{2n}{2n+1}+\frac{ 2n+1}{2n+2}\\
&=1-\frac{1}{2n+1}+1-\frac{ 1}{2n+2}\\
&=2-\frac{1}{2n+1}-\frac{ 1}{2n+2}\\
&\gt 1
\qquad\text{for }n \ge 1\\
\end{array}
$
Therefore
it can not converge.
A: The above series is divergent. It is true that it is bounded from -1 to 1 (it actually oscillates between -1 and 1), however the term "convergence" means that after a big number of steps (ergo n tends to infinity) then the series gives a unique value.
In this case it does not, since it gives 1 or -1 (if n is even or odd), therefor it is divergent. Hope that it helps you out! :)
A: The absolute value of the sequence's end behavior:
$\frac {n}{n+1}=1-\frac{1}{n+1}\to 1\neq 0 \qquad \text{as} \ \ n \to \infty$
Approaches 1, and bounded above and below by $\pm \left (1-\frac{1}{n+1}\right)$.  The sequence would then $\to (-1)^n$ as n gets larger.   Thus diverges in a similar manner to sine or cosine 
