# 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!

• Are you talking of root or ratio test? – vidyarthi Nov 9 '16 at 2:36
• Sorry! Bad typo on my end. I was talking of ratio test. I did try the root test but it ended up being inconclusive too. – Nikitau Nov 9 '16 at 2:48
• Can you rewrite the statement of the terms of the series a little more carefully. It is a little confusing to me as it stands. Also preferably put the entirety of the question into the body of the question. No important information should only be found in the title as a general rule of thumb. – jgon Nov 9 '16 at 2:53
• @Nikitau whether the power $n$ is for the whole fraction or only for 2? – vidyarthi Nov 9 '16 at 2:58
• @jgon Thanks for that! Rewrote the question and added it into the body. Should've been more careful with that. – Nikitau Nov 9 '16 at 3:12

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 $+\infty$.

• Thanks for the reply, but for some reason WolframAlpha thinks that sequence converges? Or is this an error on their end? wolframalpha.com/input/?i=does+(1%2Fn)*(2%2F(-1%5En+-3))%5En+converge – Nikitau Nov 9 '16 at 3:18
• I guess it's because you forgot to put the parenthesis around the $-1$. – André Porto Nov 9 '16 at 3:22
• great answer, avoiding the leibniz alternating test! – vidyarthi Nov 9 '16 at 3:23
• @Nikitau Be careful, on that link you sent me Wolfram did not summed the series. – André Porto Nov 9 '16 at 3:31
• Ah I see. This might be another dumb question, but which test shows that the negative terms converges? Is it by limit comparison test? – Nikitau Nov 9 '16 at 4:47

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 $\sum 1/j$.