How to prove $\sum_{n=1}^{\infty}\frac{1}{(n+1)n^{1/p}}$$\frac{1}{(n+1)n^{1/p}}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{n}{n^{1/p}}=\frac{1}{n^{1/p}}-\frac{n}{(n+1)n^{1/p}},$$
Can we go on frome here?
 A: $\sum_{n=2}^{\infty} \frac 1 {(n+1)n^{1/p}} < \int_1^{\infty} \frac 1 {(x+1)x^{1/p}} \leq \int_1^{\infty} \frac 1 {x^{1+1/p}}dx$. Evalute this integral explicitly. 
I am unable to prove the inequality when  the sum starts with $n=1$. 
A: We need to show that $$\frac{1}{(n+1)\,n^{\frac1p}}\leq p\,\left(\frac{1}{n^{\frac1p}}-\frac{1}{(n+1)^{\frac1p}}\right)$$
for every integer $n\geq 1$ and for every $p\geq 1$.  This follows from Bernoulli's Inequality
$$\left(1-\frac{1}{n+1}\right)^{\frac1p}\leq 1-\frac{1}{p}\,\left(\frac{1}{n+1}\right)\,.$$  By the way, you should have $$\sum\limits_{n=1}^\infty\,\frac{1}{(n+1)n^{\frac1p}}\leq p\,,$$ since the equality can happen when $p=1$ (and this is the only equality case).
For $0<p<1$, we have the reversed inequality:
$$\left(1-\frac{1}{n+1}\right)^{\frac{1}{p}}> 1-\frac{1}{p}\left(\frac{1}{n+1}\right)$$
for any integer $n\geq 1$.  This means
$$\sum_{n=1}^\infty\,\frac{1}{(n+1)n^{\frac1p}}>p$$
when $0<p<1$.
A: We use the classic trick of bounding the sum by an integral. Let $f(x)=1/((x+1)x^{1/p})$, then the sum is
$$\sum_{n=1}^\infty f(n)<f(1)+\int_1^\infty f(x)dx$$
(draw the graph to convince yourself of this). Of course, $x+1>x$ so that
$$f(x)=\frac{1}{(x+1)x^{1/p}}<\frac1{x\cdot x^{1/p}}=x^{-1-1/p}.$$
Therefore
$$\int_1^\infty f(x)dx<\int_1^\infty x^{-1-1/p}dx=\left.-px^{-1/p}\right]_{x=1}^\infty=p.$$
On the other hand $f(1)=1/2$, which shows that
$$\sum_{n=1}^\infty f(n)<\frac12+p.$$
Interestingly this doesn't quite prove the claim, due to that extra $+\frac12$ term. Perhaps someone else can observe how to make this bound sharper.
A: I find the proper conclusion should be
$$\sum_{n=1}^{\infty}\frac{1}{(n+1)n^{1/p}}\color{red}\leq p,$$
which is because, if $p=1$,then
$$\sum_{n=1}^{\infty}\frac{1}{(n+1)\sqrt[p]{n}}=\sum_{n=1}^{\infty}\frac{1}{(n+1){n}}=1=p.$$

Here is my proof I just figured out. First, we have
\begin{align*}
a_n:&=\frac{1}{(n+1)n^{1/p}}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{n}{n^{1/p}}=\left[\left(\frac{1}{n^{1/p}}\right)^p-\left(\frac{1}{(n+1)^{1/p}}\right)^p\right]\frac{n}{n^{1/p}}.
\end{align*}
Apply Lagrange's MVT to $f(x)=x^p$ over the interal $[1/n^{1/p},1/(n+1)^{1/p}]$. We obtain
\begin{align*}
\exists \xi \in (1/(n+1)^{1/p},1/n^{1/p})~~s.t.~~&\left(\frac{1}{n^{1/p}}\right)^p-\left(\frac{1}{(n+1)^{1/p}}\right)^p\\=&p\cdot\xi^{p-1}\cdot\left(\frac{1}{n^{1/p}}-\frac{1}{(n+1)^{1/p}}\right)\\
<&p\cdot\left(\frac{1}{n^{1/p}}\right)^{p-1}\cdot\left(\frac{1}{n^{1/p}}-\frac{1}{(n+1)^{1/p}}\right)\\
=&p\cdot\frac{n^{1/p}}{n}\cdot\left(\frac{1}{n^{1/p}}-\frac{1}{(n+1)^{1/p}}\right).
\end{align*}
Thus
$$a_n<p\left(\frac{1}{n^{1/p}}-\frac{1}{(n+1)^{1/p}}\right).$$
Therefore
\begin{align*}
\sum_{k=1}^na_k&<\sum_{k=1}^np\left(\frac{1}{k^{1/p}}-\frac{1}{(k+1)^{1/p}}\right)\\
&=p\sum_{k=1}^n\left(\frac{1}{k^{1/p}}-\frac{1}{(k+1)^{1/p}}\right)\\
&=p\left(1-\frac{1}{(n+1)^{1/p}}\right)\\
&<p.
\end{align*}
Let $n \to \infty$. It follows that
$$\sum_{k=1}^{\infty}a_k\leq p,$$
which is desired.
