Does $ \Sigma_{n=1}^{\infty} \frac{1}{n} (\frac{2}{(-1)^n - 3})^n$ converge? I'm trying to show that $ \Sigma_{n=1}^{\infty} \frac{1}{n} (\frac{2}{(-1)^n - 3})^n$ either converges or diverges. WolframAlpha is telling me it converges, but I'm not sure how to show this. I tried using the ratio test and it was inconclusive, so now I'm a little stumped on this. My definition of the root test is that if $\lim \sup \frac{c{n+1}}{c_n} < 1$, then $\Sigma c_n$ converges absolutely. The test is inconclusive if $\lim \sup \frac{c{n+1}}{c_n}  \leq \lim \inf \frac{c{n+1}}{c_n} $
Help would be very much appreciated!
 A: We have that the $2n$'th term of the sum is $\dfrac{1}{2n}$ and the $2n+1$'th term of the sum is $-\dfrac{1}{(2n+1)2^{2n+1}}$.
Then the sum of the positive terms is  $\ \displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{2n}=\dfrac{1}{2}\sum_{n=1}^{+\infty}\dfrac{1}{n}=+\infty$.
Moreover, since
$$
\left|-\dfrac{1}{(2n+1)2^{2n+1}}\right|=\dfrac{1}{(2n+1)2^{2n+1}} \leq \dfrac{1}{2^{2n+1}}
$$  the sum of the negative terms converges. Therefore, the whole sum diverges to $+\infty$.
A: There's clearly behaviour that differs between the even and odd terms that causes the use of the ratio test to be tricky.
What I'd do is combine adjacent terms, using $n=2j+1$ for odd terms and $n=2j+2$ for even. Then we can write:
$$
\sum_{n=1}^\infty \frac{1}{n}\left(\frac{2}{(-1)^n-3}\right)^n =
\sum_{j=0}^\infty \frac{1}{2j+1} \left(\frac{2}{(-1)^{2j+1}-3}\right)^{2j+1} + 
\frac{1}{2j+2} \left(\frac{2}{(-1)^{2j+2}-3}\right)^{2j+2}
$$
Some simplifying gives this as 
$$
\sum_{j=0}^\infty \frac{-1}{2j+1}\frac{1}{2^{2j+1}} + \frac{1}{2j+2}
$$
Which we can now show diverges using a limit comparison test with $\sum 1/j$.
