calculus - $sum_{k=1}^{infty}frac{1}{k(k+1)^{frac{1}{n}}}>n$

I think the following question is true:

For each positive integer $n\geq 2$, prove

$$\sum_{k=1}^{\infty}\frac{1}{k(k+1)^{\frac{1}{n}}}>n$$

I try using by induction on $n$, but I think this is not easy with induction.

Do you have any idea or comment for proving it?

So thanks for any comment and help.

Answer

The main term behaves like $k^{-\left(1+\frac{1}{n}\right)}$, hence it should not be difficult to tackle the problem through creative telescoping (section 1 here). If $a>b>0$ and $n\geq 1$ we have
$$ n(a-b)b^{n-1}\leq a^n-b^n \leq n(a-b)a^{n-1} \tag{1}$$

by simply considering the expansion of $\frac{a^n-b^n}{a-b}=a^{n-1}+\ldots+b^{n-1}$.
If we pick $a$ as $\frac{1}{k^{1/n}}$ and $b$ as $\frac{1}{(k+1)^{1/n}}$ we get:

$$\small n\left(\frac{1}{k^{1/n}}-\frac{1}{(k+1)^{1/n}}\right)\frac{1}{(k+1)^{1-1/n}}\leq \frac{1}{k(k+1)}\leq n\left(\frac{1}{k^{1/n}}-\frac{1}{(k+1)^{1/n}}\right)\frac{1}{k^{1-1/n}}$$
from which:
$$ \frac{1}{k(k+1)^{1/n}}\geq n\left(\frac{1}{k^{1/n}}-\frac{1}{(k+1)^{1/n}}\right)\tag{2} $$
leading to the claim in a straightforward way:
$$ \sum_{k\geq 1}\frac{1}{k(k+1)^{1/n}}\geq n. $$