Show that the series is convergent/divergent

$\displaystyle\sum_{n=2}^\infty = \frac{\sqrt[3]{n}\times(-1)^n}{n-1}$

I can prove it does converge, but not absolutely by using the comparison test and p-test. But to determine if it is convergent by the alternating series test, I could not prove it. I am stuck between $\sqrt[3]n$ and $\sqrt[3]{n+1}$

• You mean you cannot show that $$\frac{\sqrt[3]{n+1}}n<\frac{\sqrt[3]n}{n-1}$$? If so, cube both sides. – Simply Beautiful Art Sep 9 '17 at 13:04
• but n+1 > n while 1/n^3 <1/(n-1)^3 – Matthew Sep 9 '17 at 13:07
• You are not being clear on what your problem here is. – Simply Beautiful Art Sep 9 '17 at 13:11
• Sorry just edited it. So $\sqrt[3]{n+1}$ > $\sqrt[3]{n}$ but 1/n < 1/(n-1). – Matthew Sep 9 '17 at 13:16
• So? You need to show that $\sqrt[3]{n+1}/n<\sqrt[3]n/(n-1)$. Have you followed through on my hint by cubing both sides. – Simply Beautiful Art Sep 9 '17 at 13:19

$\sum_{n=2}^\infty(-1)^n\frac{\sqrt[3]n}{n}$ is convergent. The difference with your series is $\sum_{n=2}^\infty(-1)^n\left(\frac{\sqrt[3]n}{n} -\frac{\sqrt[3]n}{n-1}\right)$ which is absolutely convergent. So your series is convergent.
It's of course alternating and the absolute values are $\frac{\sqrt[3]{n}}{n+1}$ which is decreasing from a certain index onwards (which is enough): compute the derivative of $f(x)= \frac{\sqrt[3]{x}}{x+1}$ and check when it becomes $< 0$.
• It's clear intuitively. The $x$ increases a lot faster than the root term so order reasoning shows it must be eventually decreasing. The derivative is just a formal way to show it. @SimplyBeautifulArt – Henno Brandsma Sep 9 '17 at 13:16
• But is taking a derivative necessary for that sort of thing? Not to mention, applying derivatives and checking for $f'<0$ isn't easier than simply checking $f(x+1)<f(x)$. – Simply Beautiful Art Sep 9 '17 at 13:18