# Does this series converge conditionally $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^{\frac{1}{10}}}$

$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^{\frac{1}{10}}}$

According to my understanding, if $\sum\left|a_n\right|$ diverges but $\sum a_n$ converges, then the series is conditionally convergent.

For $\sum\left|a_n\right|$ my series can be test via the p-series test and since $\frac{1}{10} \lt 1$ it diverges.

So next I test $\sum a_n$ using the alternating series test and find that it is a decreasing series and the limit converges to 0.

Thus, I came to the conclusion that this is conditionally convergent. Is this correct?

• @user1952009 This is not what "absolutely convergent" means. Grouping the terms as one pleases is not allowed, it becomes another series. cf. Riemann's Rearrangement Theorem. – Clement C. Jul 16 '16 at 16:16
• It does change things. You are looking at another series, $\sum_n b_n$ with $b_n = a_{2n-1}+a_{2n}$. That other series is absolutely convergent, but $\sum_n a_n$ is not. – Clement C. Jul 16 '16 at 16:18
• But that statement is utterly confusing, then. What does your "it" refer to? Given that the OP is asking about a specific series, while your statement is only true if "it" refers to a different (related) series you implicitly define after using the pronoun, it can only bring confusion. (Especially since it's about a point the OP was not asking about, and most likely not considering to begin with) – Clement C. Jul 16 '16 at 16:21
• True, but then so is the (divergent, so not even well-defined) series $\sum_n (-1)^n$... :) – Clement C. Jul 16 '16 at 16:24
• (as a last comment: basically, any conditionally convergent series you can use the alternating series test on will allow the same grouping you used: pairing even and odd terms will result in an absolutely convergent series, as all terms have the same sign) – Clement C. Jul 16 '16 at 16:29

## 2 Answers

Yes, it is. (Note however that to be fully correct, the statement you make should read that "$(\lvert a_n\rvert)_n$ is a decreasing sequence," not that "$\sum_n a_n$ is a decreasing series.")

• All too easy... +1 – Mark Viola Jul 16 '16 at 19:23
• @Dr.MV "I like my proofs like I like my eggs."? – Clement C. Jul 16 '16 at 19:26
• Over easy I presume. – Mark Viola Jul 16 '16 at 19:27
• Or scrambled so nobody can refute? – Mark Viola Jul 16 '16 at 19:27
• And you would be correct. It was actually African or European. But it was only an homage, and thus need not be identical. – Mark Viola Jul 16 '16 at 19:50

A monotonically descending alternating series converges, is that series a A monotonically descending alternating series ?