# Sequence Convergence of $\sum_n\frac{(-1)^{n+1}}{3n + n(-1)^n}$

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!

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## 3 Answers

The odd terms are $a_{2k-1}=\frac1{2(2k-1)}=\frac1{4k-2}$.

The even terms are $a_{2k}=-\frac1{4(2k)}=-\frac1{8k}$

This means $$a_{2k}+a_{2k-1}=\frac1{4k-2}-\frac1{8k}=\frac{4k-(2k-1)}{8k(2k-1)}=\frac{2k+1}{8k(2k-1)}.$$

So we get for partial sums $$s_{2n}=\sum_{k=1}^n \frac{2k+1}{8k(2k-1)}\ge \sum_{k=1}^n \frac1{8k}.$$

This implies that $\lim\limits_{n\to\infty} s_{2n}=\infty$.

Since $a_{2k-1}\to 0$, we have that also $\lim\limits_{n\to\infty} s_{2n+1}=\infty$.

Together we get $$\sum_{k=1}^\infty a_n =\lim\limits_{n\to\infty} s_{n}=\infty.$$

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this is exactly what I was looking for! thank you very much martin! – Clash Nov 27 '11 at 17:49

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 $\sum\limits_{n =1}^{+\infty}u_n$.

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You know the equivalence

$$\sum_{i=1}^{n}{ \frac 1 {i}} = \log(n) + \gamma + o(1).$$

So you know that your series is equivalent to

$$-\frac 1 {4} \cdot (\log(n) + \gamma + o(1)) + \frac 1 {2}\cdot (\log(n) + \gamma + o(1)) = \frac 1 {4} \cdot (\log(n) + \gamma + o(1))$$

which does not converge.

Edit: This is a standard way to say diverges + diverges = diverges. You make equivalents until you reach $O(1)$, you add them, and you check that you have still terms that still diverge. I developed until the $o(1)$ term because, come on, it's the Euler-Mascheroni constant!

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thanks for showing this way to show that diverges + diverges = diverges! – Clash Nov 27 '11 at 18:19
To try to get an asymptotic expansion up to order $o(1)$ is a laudable goal but I think the one you got is wrong. The error lies in the step you omitted. The $n$th partial sum is $\frac12H_n-\frac38H_{n/2}$, where $H_k$ is the $k$th harmonic number, hence an equivalent is $\frac18\log(n)$ (and your constant term is wrong as well). – Did Nov 27 '11 at 18:20
I think my "mistake" lies in the OP's interpretation "n=2n" and "n=2n+1". So, yes, technically, it's false. – Fezvez Nov 27 '11 at 18:32