Hints or how to start to deduce the asymptotic behaviour of $\sum_{n\leq x}p_{n+1}^{-\frac{n}{n+1}}$, $p_k$ denotes the $k$th prime number One knows how faster diverges the sum of reciprocals of prime numbers, see for example this Wikipedia, Divergence of the sum of the reciprocals of the primes. 
For integers $n\geq 1$, we denote the $n$th prime number as $p_n$.
Thus (this reasonig is the comparison with previous statement)  after I've  assumed the Firoozbakht's conjecture, see this Wikipedia I wondered this question.

Question. In short I know as consequence of the assumption of Firoozbakht's conjecture, and since the sum of reciprocal of primes is divergent, that next sequence in $(1)$ is divergent when $x\to\infty$. What is the technique that I need to deduce (an statement about)  the asymptotic behaviour of $$\sum_{n\leq x}\frac{1}{p_{n+1}^{\frac{n}{n+1}}}\tag{1}$$
  as $x\to\infty?$ Many thanks.

Was fixed a typo in the series.
Notice that this exercise is different than (divergence and asymptotic formula  of the sum of reciprocal of primes) the explained in section 4.8 of Apostol, Introduction to Analytic Number Theory, Springer (1978). Now I need to handle the subscript $n$, that also appears in the exponent of the denominator .
 A: If we write
$$p_n^{-\frac{n-1}{n}} = \frac{1}{p_n}\cdot \sqrt[n]{p_n}$$
and note that $\sqrt[n]{p_n} \to 1$ "sufficiently fast", we note that your sum hardly differs from the sum of the reciprocals of the primes. In fact, since
$$\sqrt[n]{p_n} - 1 = \exp\biggl(\frac{1}{n}\log p_n\biggr) - 1 \sim \frac{\log p_n}{n} \sim \frac{(\log p_n)^2}{p_n}$$
it follows that
$$\sum_{n = 1}^{\infty} \frac{\sqrt[n]{p_n}-1}{p_n}$$
is convergent. Hence
\begin{align}
\sum_{n \leqslant x} p_{n+1}^{-\frac{n}{n+1}} &= -1 + \sum_{n \leqslant x+1} p_n^{-\frac{n-1}{n}} \\
&= -1 + \sum_{n \leqslant x+1} \frac{1}{p_n} + \sum_{n \leqslant x+1} \frac{\sqrt[n]{p_n} - 1}{p_n} \\
&= \log \log p_{\lfloor x\rfloor+1} + \Biggl(M-1 + \sum_{n = 1}^{\infty} \frac{\sqrt[n]{p_n}-1}{p_n}\Biggr) + O\bigl((\log x)^{-1}\bigr) - \sum_{n > x+1} \frac{\sqrt[n]{p_n}-1}{p_n} \\
&= \log \log x+ \Biggl(M-1 + \sum_{n = 1}^{\infty} \frac{\sqrt[n]{p_n}-1}{p_n}\Biggr) + o(1).
\end{align}
We can estimate the tail sum
$$\sum_{n > x+1} \frac{\sqrt[n]{p_n}-1}{p_n} \in O\biggl(\frac{(\log x)^2}{x}\biggr)$$
easily, and
$$\log \log p_k = \log (\log k + O(\log \log k)) = \log \log k + O\biggl(\frac{\log \log k}{\log k}\biggr)$$
shows that our $o(1)$ is in fact $O\bigl(\frac{\log \log x}{\log x}\bigr)$.
