Testing the convergence of difference of cube roots

Question

I'm studying for my exams and I came across some problems for which I can't find any kind of example:

$$\sum_{n=0}^\infty \sqrt[3]{n+1} - \sqrt[3]{n} $$

and then:

$$\sum_{n=0}^\infty \sqrt[3]{n^2+1} - \sqrt[3]{n^2} $$

so obviously there is something I should learn to do here. After that, there is the same problem but with the 5th root.

If someone could help me solve it I would be more than thankful.

Answer 1

Hint:$$\sqrt[3]{n+1}-\sqrt[3]n=\frac1{\sqrt[3]{n+1}^2+\sqrt[3]{n+1}\sqrt[3]n+\sqrt[3]n^2}.$$

Answer 2

$\sum_{n=0}^N \sqrt[3]{n+1} - \sqrt[3]{n}$ is just $\sqrt[3]{N+1}-\sqrt[3] 0$, because intermediate terms cancel. This tends to infinity, so the sum diverges.

Answer 3

For the first question, note that you have a telescoping sum, so you can write down explicitly the partial sums $$ \sum_{n=0}^N \left(\sqrt[3]{n+1}-\sqrt[3]n\right)=\sqrt[3]{N+1}-\sqrt[3]{0}=\sqrt[3]{N+1} $$ which $\to\infty$ as $N\to\infty$.

For the second sum, estimate $$ \sqrt[3]{n^2+1}-\sqrt[3]{n^2}=\frac{(n^2+1)-n^2}{(\sqrt[3]{n^2+1})^2+\sqrt[3]{n^2}\sqrt[3]{n^2+1}+(\sqrt[3]{n^2})^2}\sim \frac1{3n^{4/3}} $$ and so converges.