Consider the series
![\[ \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{\frac{1}{n}}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f3c943d5a6b1af9a1b8567795513475e_l3.png "Rendered by QuickLaTeX.com")
Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.
The given series diverges.
Proof. The series diverges since the limit
![\[ \lim_{n \to \infty} \frac{(-1)^n}{n^{\frac{1}{n}}} \neq 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e172dba29697977b62b3e4a3f0b820ce_l3.png "Rendered by QuickLaTeX.com")
(Since the limit of the even terms is 1 and the limit of the odd terms is 
If Leibniz’s rule fails, we can not conclude divergence.
If Leibniz’s rule fails, we can not conclude divergence.
Why are you considering the (-1)^{n} when taking the limit? Shouldn’t you just consider a_n=\cfrac {1}{n^{\cfrac{1}{n}}}?
That’s what I thought, though I think you can use theorem 10.6 since there’s no mention about the positivity of “an”, and in this case “an” goes to -1 or 1 (not zero)