# 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 .

• At first only need to know how to start the exercise, but feel free if you want to discuss some detail, notice that with the purpose to create this question I've assumed Firoozbakht's conjecture, but I would like to know unconditionally an statement (a very good deduction is not required) about the asymptotic behaviour of $(1)$ to finish the comparison with the divergence of the sum of reciprocal of prime numbers. I did comparisons using Wolfram Alpha online calculator.
– user243301
Nov 22 '17 at 13:11
• The first step is independent of Firoozbakht's conjecture. If $a_k \geqslant 0$, $\sum a_k$ diverges, and $b_k \to 1$, then $$\sum_{k = 1}^n a_kb_k \sim \sum_{k = 1}^n a_k\,.$$ Nov 22 '17 at 13:49
• Mertens' theorems give $\sum_{p\leq x}\frac{1}{p}\approx\log\log x$ so it is reasonable to expect that your sum behaves like $C\log\log x$ with $C$ being a constant close to $1$. The way to go is summation by parts, for sure. Nov 22 '17 at 18:50
• Once for all : if the sum diverge and $f$ is non-negative and nice enough then $\sum_{n \le x} f(p_n) \sim \sum_{k \le x \log x} f(k \log k)$ and $\sum_{p\le x} f(p) \sim \sum_{n \le x} \frac{f(n)}{\log n}$, this is the prime number theorem. Asking 100 times the same questions won't change the answer. Nov 23 '17 at 0:19
• What about the case $\sup_n f(p_n)$ converges ? Nov 23 '17 at 8:02

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 $O\bigl(\frac{\log \log x}{\log x}\bigr)$.

• Sure that now I am going to study your answer. Thanks!!!
– user243301
Nov 22 '17 at 20:33
• Can you refer (I know that you've splitted the infinite series) where the term $O\bigl((\log x)^{-1}\bigr)$ comes from? Is it Mertens theorem? And how do you deduce (what is the reasoning/property of logarithms that you're invoking) the term $O\biggl(\frac{\log \log k}{\log k}\biggr)$ from your last identity $\log (\log k + O(\log \log k)) = \log \log k + O\biggl(\frac{\log \log k}{\log k}\biggr)$? Many thanks.
– user243301
Nov 22 '17 at 20:46
• The $O\bigl((\log x)^{-1}\bigr)$ comes from Mertens' theorem, using $\log p_k \sim \log k$. The last is from \begin{aligned}\log (\log k + O(\log \log k)) &= \log \biggl((\log k)\biggl(1 + \frac{O(\log \log k)}{\log k}\biggr)\biggr) \\&= \log \log k + \log \biggl(1 + \frac{O(\log \log k)}{\log k}\biggr) \\&= \log \log k + \frac{O(\log \log k)}{\log k}.\end{aligned} Nov 22 '17 at 20:51
• Just I was thinking and I've edited my comment a second ago, many thanks for your patience and all these details.
– user243301
Nov 22 '17 at 20:52