# 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. Commented Jul 16, 2016 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. Commented Jul 16, 2016 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) Commented Jul 16, 2016 at 16:21
• True, but then so is the (divergent, so not even well-defined) series $\sum_n (-1)^n$... :) Commented Jul 16, 2016 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) Commented Jul 16, 2016 at 16:29

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.")