# $\sum{\frac{(-1)^{n-1}}{{(n(n+1))^{1/3}}}}$ series convergence

$$\sum{\frac{(-1)^{n-1}}{{(n(n+1))^{1/3}}}}$$

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.

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

For absolute convergence:

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

$$\sum\frac1{(n(n+1))^{1/3}}=\sum\frac1{(n^2+n)^{1/3}}=\sum\frac1{n^{2/3}(1+\frac1n)^{1/3}}$$

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 $(-1)^n.$

• 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. – Sri-Amirthan Theivendran Jun 24 '17 at 15:27
• @FoobazJohn Thank you for spotting that. I've corrected the answer. – Sahiba Arora Jun 24 '17 at 15:33
• @iskra Please note the corrections. – 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 $n \to \infty$, $$\frac{1}{{(n(n+1))^{1/3}}} \to 0.$$

• 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? – mathbbandstuff Jun 24 '17 at 14:46
• @iskra substitute $n\mapsto n+1$ and $\sum \frac{(-1)^n}{((n+1)(n+2))^{1/3}}$, which the above test is still $\sim \frac{1}{n^{2/3}}$ – 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.