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?

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.
