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

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 $\sqrt[3]{n+1}$

  • $\begingroup$ You mean you cannot show that $$\frac{\sqrt[3]{n+1}}n<\frac{\sqrt[3]n}{n-1}$$? If so, cube both sides. $\endgroup$ – Simply Beautiful Art Sep 9 '17 at 13:04
  • $\begingroup$ but n+1 > n while 1/n^3 <1/(n-1)^3 $\endgroup$ – Matthew Sep 9 '17 at 13:07
  • $\begingroup$ You are not being clear on what your problem here is. $\endgroup$ – Simply Beautiful Art Sep 9 '17 at 13:11
  • $\begingroup$ Sorry just edited it. So $\sqrt[3]{n+1}$ > $\sqrt[3]{n}$ but 1/n < 1/(n-1). $\endgroup$ – Matthew Sep 9 '17 at 13:16
  • $\begingroup$ 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. $\endgroup$ – 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 $< 0$.

  • $\begingroup$ Using derivatives is somewhat overkill? $\endgroup$ – Simply Beautiful Art Sep 9 '17 at 13:14
  • $\begingroup$ 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 $\endgroup$ – Henno Brandsma Sep 9 '17 at 13:16
  • $\begingroup$ 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 $f(x+1)<f(x)$. $\endgroup$ – Simply Beautiful Art Sep 9 '17 at 13:18

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