# From an attempt to explore relationships between the AM-GM inequality and prime conjectures, to the asymptotic of certain arithmetic functions

We denote the $k$th prime number as $p_k$, and if it is required in your calculations let $g_k=p_{k+1}-p_k$ denoting the gaps between primes.

For integers $n\geq 1$, I define the Andrica's function as the arithmetic function (I don't know if this definition and next claim was in the literature; at first isn't related to Andrica's conjecture but I was inspired in it, see this MathWorld) $$\mathcal{A}(n)=\sqrt[n]{\prod_{k=1}^n\left(\sqrt{p_{k+1}}-\sqrt{p_{k}}\right)}.\tag{1}$$

Claim. Then $$\lim_{n\to\infty}\mathcal{A}(n)=0,$$ and thus $$\mathcal{A}(n)=o(1),$$ as a consequence of a combination of the AM-GM inequality with the prime number theorem and the squeeze theorem.

On the other hand I define the Firoozbakht's function as the function defined for integers $n\geq 1$ $$\mathcal{F}(n)=\sqrt[n]{\prod_{k=1}^n p_{k}\left(p_k^{1/k}-1\right)}.\tag{2}$$ (This definition is inspired in Firoozbakht's conjecture, see this Wikipedia.)

Question.

A) (Optional) Can you improve my knowledge of the function defined in $(1)$ with regard to its asymptotic behaviour as $n\to\infty$? Then provide me a better statement than my Claim about the asymptotic behaviour of $\mathcal{A}(n)$ as $n\to\infty$? You can add your statement as a big/little oh statement or as an asymptotic equivalence. If it is a well-known exercise, then refers the literature and I try to search and read those calculations.

B) Hint or details to get a statement about the asymptotic behaviour of $\mathcal{F}(n)$ as $n\to\infty$. Many thanks.

It will be welcome also if you want to add to your calculations some conjecture or conditional statement (on assumption of well-known conjectures), or well numeric computations or graphics for $1\leq n\leq N$ involving our functions $\mathcal{A}(n)$ and $\mathcal{F(n)}$ for a large $N$.

• isn't $\mathcal{F}(n)$ always larger than the geometric mean of the first $n$ primes, since $p_k^{1/k}>1$? – Mastrem Dec 2 '17 at 15:40
• Yes, if there were not mistakes in my calculations I've deduced the same thing, but I think that it is not a very well deduction (since the productmany times of an inequality maybe isn't a good statement). Any case many thanks @Mastrem I hope understand your previous words. – user243301 Dec 2 '17 at 19:18
• ${\cal F}(n)$ is not larger than the geometric mean of the first $n$ primes because $p_k^{1/k}-1 \to 0$. See my answer below for a more realistic estimate of the order of magnitude of ${\cal F}(n)$. – Alex Dec 5 '17 at 4:41

We have (by Theorem 5 in arXiv:1506.03042) $$p_k^{1+1/k} - p_k = \log^2 p_k - \log p_k - 1 + o(1).$$ Therefore $${\cal F}(n) = O(\log^2 p_n) = O(\log^2 n).$$