Does this converge or diverge (absolute and/or conditional)?$\\$

I've tried Leibniz, D'Alembert, Cauchy and Cauchy-integral criteria, all that's left is the comparison test (those are the only things I can use). Any hints would be helpful.

  • 2
    $\begingroup$ Leibniz${}{}{}{}$ $\endgroup$ – kingW3 Jun 24 '17 at 14:38
  • $\begingroup$ It isn't absolutely convergent as $\sum \frac{1}{(n(n+1))^{1/3}} \approx \sum \frac{1}{n^{2/3}}$ $\endgroup$ – Sahiba Arora Jun 24 '17 at 14:45
  • $\begingroup$ @SahibaArora I thought about that too, but I'm not sure how to precisely prove it.. $\endgroup$ – mathbbandstuff Jun 24 '17 at 14:47
  • 1
    $\begingroup$ @iskra I have added an answer to show it precisely. $\endgroup$ – Sahiba Arora Jun 24 '17 at 15:04

For absolute convergence:

Let $a_n=\frac1{(n(n+1))^{1/3}}$. Now,


Let $b_n=\frac1{n^{2/3}}$ (divergent) $$\lim \frac{a_n}{b_n}=\lim \frac1{(1+\frac1n)^{1/3}}=1\text{ (finite and non-zero)}$$

By Limit Comparison Test, the series is not absolutely convergent.

For conditional convergence:

It is easy to see that $\frac{1}{(n(n+1))^{1/3}}$ is monotonically decreasing and converges to $0$. So by Leibniz's Test it is convergent.

Note: Leibniz test says that $\sum (-1)^na_n$ converges. But you can write this as $\sum(-1)^{n-1}a_{n-1}$. Therefore, it doesn't make a difference that you have $(-1)^{n-1}$ in your series instead of $(-1)^n.$

  • $\begingroup$ Your first inequality isn't strong enough. It says that the series in question is bounded by $\infty$ which doesn't prove that it diverges. $\endgroup$ – Sri-Amirthan Theivendran Jun 24 '17 at 15:27
  • $\begingroup$ @FoobazJohn Thank you for spotting that. I've corrected the answer. $\endgroup$ – Sahiba Arora Jun 24 '17 at 15:33
  • $\begingroup$ @iskra Please note the corrections. $\endgroup$ – Sahiba Arora Jun 24 '17 at 15:33

One may use the alternating series test, just check that $$ n \mapsto \frac{1}{{(n(n+1))^{1/3}}} $$ is monotonically decreasing over $[1,\infty)$ and observe that, as $n \to \infty$, $$ \frac{1}{{(n(n+1))^{1/3}}} \to 0. $$

  • $\begingroup$ But Leibniz would give the convergence of $(-1)^n \cdot a_n$ and in our expression we have $(-1)^{n-1}$ hence our $a_n$ is supposed to be $\frac{-1}{(n(n+1))^{1/3}}$ which is monotonically increasing...also what about absolute convergence? $\endgroup$ – mathbbandstuff Jun 24 '17 at 14:46
  • 1
    $\begingroup$ @iskra substitute $n\mapsto n+1$ and $\sum \frac{(-1)^n}{((n+1)(n+2))^{1/3}}$, which the above test is still $\sim \frac{1}{n^{2/3}}$ $\endgroup$ – Dando18 Jun 24 '17 at 15:03

Leibniz's Test is fine here: the series is alternating and $1/(n(n+1))^{1/3}$ decreases to zero. Hence the series is convergent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.