Does the series $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{3n+n(-1)^n}$ converge?

Question

I have to find out if the series $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{3n+n(-1)^n}$$ converges. Root test and ratio test did not work out for me. I also tried the alternating series test, but I can not use it because of the $${(-1)^{n+1}}$$ in the denominator. Also, I have to find the limit of the series if it does converge.

Answer 1

By observation, the series diverges. In order to prove it diverges, it is sufficient to prove it by proving the grouped series diverges. $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{3n+n(-1)^n}\text{ diverges if $\sum_{n=1}^\infty \frac{1}{3(2n-1)-(2n-1)}-\frac{1}{3\cdot 2n+2n}$diverges.}\\ $$But we have$$\sum_{n=1}^\infty \frac{1}{3(2n-1)-(2n-1)}-\frac{1}{3\cdot 2n+2n}\\ =\sum_{n=1}^\infty \frac{2 n+1}{8 n (2 n-1)}=+\infty$$

Answer 2

It diverges because $$\sum_{n=1}^{\infty}\frac{-1}{n\left(1+3\left(-1\right)^{n}\right)}=\sum_{n=1}^{\infty}\left(\frac{1}{2\left(2n-1\right)}-\frac{1}{8n}\right)=\sum_{n=1}^{\infty}\frac{2n+1}{8n\left(2n-1\right)}$$ obtain $O(\frac{1}{n})$ terms.