<s>[INST] Problem with showing convergence or divergence of $\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$ Determine if the following series is absolutely convergent, conditionally convergent or divergent.
$$\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$$
So I showed that$\sum_{n=2}^\infty\bigl|\frac{(-1)^{n+1}}{3n+4}\bigr|$ is divergent using limit comparison test.
Now for $\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$, I considered 2 ways:


*

*Alternating Series Test
$$a_n:=\frac{1}{3n+4}$$
$$3n+4>0\implies \frac{1}{3n+4}>0\implies a_n\ge0$$
$$\lim_{n\to \infty} a_n=\lim_{n\to \infty}\frac{1}{3n+4}=\lim_{n\to \infty}\frac{\frac{1}{n}}{3+\frac{4}{n}}=\frac{0}{3+0}=0$$
$$3n+4\le 3n+7\implies a_n=\frac{1}{3n+4}\ge \frac{1}{3n+7}=a_{n+1}$$
Therefore $\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$ is convergent.

*Comparison Test (Note: $3n+4\ge0$)
$$-1\le(-1)^{n+1}\le 1 \implies \frac{-1}{3n+4} \le \frac{(-1)^{n+1}}{3n+4} \le \frac{1}{3n+4}$$
But I have already showed that $\sum_{n=2}^\infty\frac{1}{3n+4}$ diverges.
and consequently, $-\sum_{n=2}^\infty\frac{1}{3n+4}$ diverges as well. So that would imply that $\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$ is divergent!
Did i make a computational error somewhere? I can't seem to find it.
 [/INST] The alternating test is OK.
Now, there is no contradition with your second point, because
$$
\frac{-1}{3n+4} \le \frac{(-1)^{n+1}}{3n+4} \le \frac{1}{3n+4}
$$ yields
$$
-\infty \le \sum_{n\geq1}\frac{(-1)^{n+1}}{3n+4} \le +\infty
$$ which does not exclude convergence of the initial series.
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