TITLE: Sequence Convergence of $\sum_n\frac{(-1)^{n+1}}{3n + n(-1)^n}$
QUESTION [6 upvotes]: I have the following series $\displaystyle \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{3n + n(-1)^n}$. Does it converge? 
I wanted to the alternating series test, but that's not easy because of the two $(-1)^n$, plus it's not monotone decreasing.
I've split it into two series
$n=2n \Rightarrow \sum{-\frac 1 {4m}}\Rightarrow$ diverges,
$n=2n+1 \Rightarrow \sum{\frac 1 {2m}}\Rightarrow$ diverges.
I can't say diverges + diverges = diverges, but I have this hunch, that the positive series diverges faster than the negative series and therefore it diverges. However, I don't know any test I can do to prove that? Alternating, ratio and root fails. I don't know with what I can compare it to. Thanks in advance for your help!

REPLY [3 votes]: We know by the alternating series test that the series $\sum\limits_{n =1}^{+\infty}\frac{(-1)^{n+1}}{3n}$ is convergent. Therefore, the series $\sum\limits_{n =1}^{+\infty}u_n$ is convergent if and only if $\sum\limits_{n =1}^{+\infty}\left(u_n-\frac{(-1)^{n+1}}{3n}\right)$ converges. With $u_n=\frac{(-1)^{n+1}}{3n+n(-1)^n}$, we get 
$$u_n-\frac{(-1)^{n+1}}{3n}=\frac{(-1)^{n+1}}n\left(\frac 1{3+(-1)^n}-\frac 13\right)=\frac{(-1)^{n+1}}n\frac{3-(3+(-1)^n)}{3+(-1)^n}
$$
hence 
$$u_n-\frac{(-1)^{n+1}}{3n}=\frac{(-1)^{n+1}(-1)^n(-1)}{n(3+(-1)^n)}=\frac 1{3n+n(-1)^n}\geq\frac 1{4n},$$
which shows that $\sum\limits_{n =1}^{+\infty}u_n-\frac{(-1)^{n+1}}{3n}$ is divergent, hence so is $\sum\limits_{n =1}^{+\infty}u_n$.