Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 8 down vote accepted

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

share|cite|improve this answer
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 $\sum\limits_{n =1}^{+\infty}u_n$.

share|cite|improve this answer

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!

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.