Infinite series inequality. I have some trouble with the following problem:
Show that $$\sum_{n=1}^{\infty} \frac{1}{(n+1)\sqrt[p]{n}}<p$$ is true for every natural number $p$. 
I tried using induction but I can't finish the proof. Or there is a simpler way to solve this? Thank you.

Solution 1. Let $a = \sqrt[p]{n}$ and $b = \sqrt[p]{n+1}$. If $p > 1$ is an integer, then
\begin{align*}
\frac{1}{(n+1)\sqrt[p]{n}}
&= \frac{p}{ab \cdot pb^{p-1}} \\
&< \frac{p}{ab \cdot (b^{p-1} + ab^{p-2} + \cdots + a^{p-1})} \\
&= p \cdot \frac{b - a}{ab (b^p - a^p)} \\
&= p \left( \frac{1}{\sqrt[p]{n}} - \frac{1}{\sqrt[p]{n+1}} \right).
\end{align*}
Summing over $n = 1, 2, \cdots$ gives the desired inequality.

Solution 2. Let $p \geq 1$ be real and write
$$ p \left( \frac{1}{n^{1/p}} - \frac{1}{(n+1)^{1/p}} \right)
= \frac{p}{(n+1)n^{1/p}} \left[ (n+1) - n^{\frac{1}{p}}(n+1)^{1-\frac{1}{p}} \right]. $$
By the Jensen's inequality (or by the AM-GM inequality if $p$ is integer), we have
$$n^{\frac{1}{p}}(n+1)^{1-\frac{1}{p}} \leq \frac{1}{p}\cdot n + \left(1-\frac{1}{p}\right)\cdot(n+1) = n+1 - \frac{1}{p} $$
with the equality exactly when $p = 1$. So it follows that
$$ p \left( \frac{1}{n^{1/p}} - \frac{1}{(n+1)^{1/p}} \right) \geq \frac{1}{(n+1)n^{1/p}} $$
with equality if and only if $p = 1$. Summing over $n = 1, 2, \cdots $ gives the desired inequality.
