# 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:

1. 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.

2. 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.

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.