$\sum_{k=1}^{\infty}\frac{1}{k(k+1)^{\frac{1}{n}}}>n$ 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.
 A: This solution works for any real number $n\geq 1$.  Using Bernoulli's Inequality, we have 
$$
\begin{align}\frac{n}{k^{1/n}}-\frac{n}{(k+1)^{1/n}}&=\frac{n}{(k+1)^{1/n}}\,\left(\left(1+\frac{1}{k}\right)^{1/n}-1\right)
\\
&\leq\frac{n}{(k+1)^{1/n}}\,\Biggl(\left(1+\frac{1}{nk}\right)-1\Biggr)=\frac{1}{k(k+1)^{1/n}}\end{align}$$
for every positive integer $k$.  The equality holds iff $n=1$.
On the other hand, we can see that
$$\begin{align}
\frac{1}{k(k+1)^{1/n}}&<\frac{1}{k^{1+1/n}}=\frac{n}{k^{1/n}}\Biggl(\left(1+\frac{1}{nk}\right)-1\Biggr)
\\
&\leq\frac{n}{k^{1/n}}\Biggl(\left(1-\frac{1}{k}\right)^{-1/n}-1\Biggr)=\frac{n}{(k-1)^{1/n}}-\frac{n}{k^{1/n}}
\end{align}$$
for every integer $k\geq 2$.  This proves that
$$n\leq \sum_{k=1}^\infty\,\frac{1}{k(k+1)^{1/n}} <\frac{n+1}{2^{1/n}}<n+1-\ln(2)<n+\frac{1}{2}\,.$$
The left-hand side of the inequalities above becomes an equality if and only if $n=1$.  If $0<n<1$, we have a different bound:
$$\frac{n}{2^{1/n}}<\sum_{k=1}^\infty\,\frac{1}{k(k+1)^{1/n}}<n\,.$$

Interestingly, a similar argument yields the inequality $$n-1<n-\ln(2)<\frac{n}{2^{1/n}}<\sum_{k=1}^\infty\,\frac{1}{k^{1/n}(k+1)}\leq n$$ for every real number $n\geq 1$.  The right-hand side of the inequalities above becomes an equality if and only if $n=1$.  If $0<n<1$, we have a different bound: $$\frac{n}{2^{1/n}}+\frac{1}{2}<\sum_{k=1}^\infty\,\frac{1}{k^{1/n}(k+1)}<n+\frac{1}{2}\,.$$

A: 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:
$$ \sum_{k\geq 1}\frac{1}{k(k+1)^{1/n}}\geq n. $$
A: Hint
It doesn't completely prove the result, but a rough lower bound can be obtained by
$$\frac{1}{k(k+1)^{1/n}}\ge\frac{1}{(k+1)^{1+\frac{1}{n}}}\ge \int_{k+1}^{k+2}\frac{1}{x^{1+\frac{1}{n}}} d x$$
A: Since $\zeta(s)>\dfrac{1}{s-1}$ for $s>1$ then
$$\sum_{k=1}^{\infty}\frac{1}{k(k+1)^{\frac{1}{n}}}>\sum_{k=1}^{\infty}\frac{1}{(k+1)^{1+\frac{1}{n}}}=\zeta\left(1+\frac{1}{n}\right)-1>n-1$$
