Is the series convergent $\sum (n+1)^{\frac{1}{3}} - n^{\frac{1}{3}}$

Question

$\sum (n+1)^{\frac{1}{3}} - n^{\frac{1}{3}}$ here $n \to \infty$

Now, I tried to do this by comparison as:

$$n^{\frac{1}{3}} ([ 1+ \frac{1}{n}]^{\frac{1}{3}}-1) < n^{\frac{1}{3}}$$

now, from p series we know that $$n^{\frac{1}{3}}$$ is divergent, but if the series we compared with the given equation is divergent is > then it is not necessary that the given series is divergent (As far as I know).

And I can't find any other way to solve this.

Answer 1

Note that as $n\to +\infty$, $$(n+1)^{\frac{1}{3}} - n^{\frac{1}{3}}=n^{1/3}\left(\left(1+\frac{1}{n}\right)^{\frac{1}{3}} - 1\right)\sim n^{1/3}\cdot \frac{1}{3n}\sim \frac{1}{3n^{2/3}}$$ so the series is divergent because $2/3\leq 1$

Answer 2

This is a telescoping series hence the results is $\lim_{n} (\sqrt[3]{n}-1)=\infty$, hence the series diverge.

Answer 3

No. You can use Cauchy intermediate value theorem, to get $(n+1)^{\frac 1 3}-n^{\frac 1 3}= \frac 1 3 x^{-\frac 2 3}$, for some $x\in(n,n+1)$. Since $f(x)= \frac 1 3 x^{-\frac 2 3}$ is decreasing, $(n+1)^{\frac 1 3}-n^{\frac 1 3}> \frac 1 3 (n+1)^{-\frac 2 3}$. Therefore, $\sum (n+1)^{\frac 1 3}-n^{\frac 1 3} > \frac 1 3 (n+1)^{-\frac 2 3}=\infty$.

Answer 4

One way is: $$(n+1)^\frac13-n^\frac13=\dfrac{1}{\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}+\sqrt[3]{n^2}}>\dfrac{1}{3(n+1)}$$