Prove that $x_n = (-1)^n \frac {3n + 2} {n + 1}$ is divergent Prove that the sequence {${x_n}$} to be defined by $x_n = (-1)^n \frac {3n + 2} {n + 1}$ is divergent.
Also, you can assume without proof that $\lim{n\to\infty}$ $\frac{3n+2}{n+1} = 3 $
Proposed Solution: (by contradiction)
Given that there were a limit $x$, then for $\varepsilon = \frac{1}{2}$ we expect an $M$ to satisfy the divergence definition. Suppose such an $M$ exists. Then for $n\geq M$ we need to compute:
$\frac{1}{2} > |x_n - x| = |1 - x|$ and $\frac{1}{2} > |x_{n+1} - x| = |-1 - x|$

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*Do I do the following next? : plug in $x_{2n} = \frac{6n}{2n+1}$ and $x_{2n+1} = -\frac{6n+3}{2n+2} $


*Then conclude this? : $\lim{n\to\infty}$ $x_{2n} = 3$ and $\lim{n\to\infty}$ $x_{2n+1} = -3$


*Therefore, $\frac{1}{2} \ngtr |1 - 3|$ and $\frac{1}{2} \ngtr |-1 - (-3) |$ ?

Suppose the limit exists and call it $x$. Thus, for $\epsilon = \dfrac{1}{2}$, there is a $n$ such that $|x_n - x| < \dfrac{1}{2}, |x_{n+1} - x| < \dfrac{1}{2}\implies |x_{n+1} - x_n| < 1$. If $n$ is even, then $|x_n - x_{n+1}| = \left|\dfrac{3n+2}{n+1} + \dfrac{3n+5}{n+2}\right| > 2$, contradiction. The case $n$ is odd is similar. So $x$ can't exist.
