# Prove divergence of $\sum\limits_{n=1}^\infty\left(\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right)^{2/3}$

Prove divergence of $$\sum\limits_{n=1}^\infty\left(\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right)^{2/3}$$.

I need to prove that this series is divergent, but the ratio test, root test and the divergent majorant $$\frac{1}{n}$$ are inconclusive, so how would I be able to prove this?

I also tried it being proportional to something, but I couldn't get a divergent majorant out of it, because for large n it always becomes smaller.

• $$\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)^{2/3}=\left(\frac{1}{\sqrt{n}\,\sqrt{n+1}\,(\sqrt{n+1}+\sqrt n)}\right)^{2/3}\sim\left(\frac{1}{2n^{3/2}}\right)^{2/3}=\cdots$$ – Did Jan 23 '19 at 9:33
• @Did In your opinion is "This question is missing context or other details"? – Robert Z Jan 23 '19 at 9:47
• – Martin R Jan 23 '19 at 9:50
• @RobertZ It is borderline, on the one hand the OP mentions some tries, on the other hand some declarations are suspect, for example " I couldn't get a divergent majorant out of it, because for large n it always becomes smaller". Why do you ask? – Did Jan 23 '19 at 9:53
• @Did Because a few weeks ago I answered to a very similar "borderline" question and then the question has been quickly deleted. I wonder what is going happen to this question... – Robert Z Jan 23 '19 at 10:00

Note that $$\frac 1{\sqrt n} - \frac 1{\sqrt {n+1}} = \frac {\sqrt{n+1} - \sqrt n}{\sqrt {(n+1)n}} = \frac 1 {\sqrt {n(n+1)} \cdot (\sqrt n + \sqrt{n+1})} \sim \frac 1{2n^{3/2}} \quad [n \to \infty]$$ hence the term is asymptotically $$(1/2)^{2/3} n^{-1}$$, and the series diverges.
$$\frac 1 {\sqrt n} -\frac 1 {\sqrt {n+1}} =\frac 1 {\sqrt n \sqrt {n+1}(\sqrt n +\sqrt {n+1})}$$. Compare the given series with $$\sum \frac 1 n$$ which is divergent.
$$\left(\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right)^{2/3}=\left(\frac{\sqrt{n+1}-\sqrt n}{\sqrt n\sqrt{n+1}}\right)^{2/3}=\left(\frac{1}{\sqrt n\sqrt{n+1}(\sqrt{n+1}+\sqrt n)}\right)^{2/3}>\left(\frac{1}{\sqrt{n+1}\sqrt{n+1}(\sqrt{n+1}+\sqrt{n+1})}\right)^{2/3}=\frac{2^{-2/3}}{n+1}$$