Series divergent

Question

I have a problem with proof of the divergence of this series:

$$\ \sum_{i=1}^\infty \frac{\sqrt {n+1} - \sqrt n }{\sqrt[3]n}$$

I got: $$\ \frac{1}{\sqrt[3]n(\sqrt {n+1} + \sqrt n)} \to 0$$

However, how can I prove that it is divergent? I suppose I have to use the comparison test.

Answer 1

$\dfrac{1}{n^{1/3}(\sqrt{n+1}+√n)}\gt$

$\dfrac{1}{n^{1/3}2\sqrt{n+1}}\gt$

$\dfrac{1}{2n^{1/3}\sqrt{n+n}}=$

$\dfrac{1}{2√2n^{1/3}n^{1/2}}=$

$\dfrac{1}{2√2 n^{5/6}} $;

$\sum \dfrac{1}{n^p}$ diverges for $p \le 1$.

Answer 2

Since\begin{align}\lim_{n\to\infty}\frac{\frac{\sqrt{n+1}-\sqrt n}{\sqrt[3]n}}{\frac1{n^{5/6}}}&=\lim_{n\to\infty}\frac{n^{5/6}}{\sqrt[3]n\left(\sqrt{n+1}+\sqrt n\right)}\\&=\frac12,\end{align}and the series $\sum_{n=1}^\infty\frac1{n^{5/6}}$ diverges, your series diverges.