# Sequence with Prime Numbers

I was looking a question in a calculus book which used the following steps to show that following sequence has a limit (called Euler's constant $\gamma$):

$$t_n = \sum_{i=1}^n\left(\frac{1}{n}\right) -\log(n).\tag 1$$

The book wants the reader to interpret the following as difference of area:

$$t_n - t_{n+1} = [\log(n+1) - \log(n)] - \frac{1}{n+1}. \tag 2$$

Here is where things change with me. I also was looking at a rewrite of the Firoozbakht's conjecture:

$$\frac{\log(p_{n+1})}{\log(p_n)} < \frac{n+1}{n}. \tag 3$$

I noticed that what is inside the [] brackets of $(2)$ is the log of the right side of $(3)$. So, I wondered if there is a constant with the difference of area with the loglog of consecutive primes

$$a_n - a_{n+1} = [\log\log{(p_{n+1})} - \log\log{(p_n)}] - \frac{1}{f(n+1)}. \tag 4$$

Where $f(n+1)$ is the $n+1$ term of some function in which I am unsure of at the moment. But this would lead to a sequence $a_n$

$$a_n = \sum_{i=1}^n\left(\frac{1}{f(n)}\right) -\log\log(p_n), \tag 5$$ and
$$\lim_{n\to\infty} a_n = C.$$

I started to write some questions with this when I read about the Meissel–Mertens constant $$M := \lim_{k \to \infty} \left\{ \left( \sum_{i=1}^{k} \frac{1}{p_i} \right) - \log \log p_k \right\} = \lim_{n \rightarrow \infty } \left\{ \left( \sum_{p \leq n} \frac{1}{p} \right) - \log \log n \right\}. \,$$

And, I found this paper: https://projecteuclid.org/euclid.pja/1296570390. The generator in this paper for this sequence answers one of my questions. He uses $B$ instead of $M$ also.

So, the questions I still have are: What are valid functions for $f(n)$ other than $p_n$? And, what other value can the constant $C$ hold? Does the constant $M$ and/or $\gamma$ imply anything with the Firoozbakht's conjecture? What would happen when using different function like using $f(n) := n$ or some other value (like $p_n$), in other words what ranges work? Does $M - \gamma$ mean anything?

The following has been added since the question was asked. As to the last question:

Let $M := \lim_{k \to \infty} \left\{\sum_{i=1}^{k}\left(\frac{1}{p_i}\right) - \log \log p_k \right\}$, and $\gamma := \lim_{n \to \infty} \left\{\sum_{i=1}^n\left(\frac{1}{n}\right) -\log(n)\right\}$. We know that both of the constants and there limits well, and we know that: $$\lim_{n\to\infty} (a_n + b_n) = \lim_{n\to\infty} a_n + \lim_{n\to\infty} b_n.$$ So, we can combine these into: $$M - \gamma := \lim_{n\to\infty} s_n = \lim_{n \to \infty} \left\{ \sum_{i=1}^{n}\left(\frac{1}{p_i} - \frac{1}{i}\right) - F_n\right\} \text{, with } F_n = \log\log{p_n} - \log{n}.\tag 1$$

Now with $F_n$ and take the exp twice: $$\frac{\log(p_n)}{n} = e^{F_n}$$ $$p_n^\frac{1}{n} = e^{e^{F_n}}$$ The LHS is the $n$-th term of the Firoozbakht's conjecture sequence $a_n = p_n^\frac{1}{n}$. Now, because of this, what does this show in respect to the prior questions?

• I was wondering when you'd get around to asking this question! Unfortunately, the answer to the 'do these constants imply anything about the conjecture' is no; the Euler and Meissel-Mertens formulae are concerned with the asymptotic behavior of a sequence, whereas the Firoozbakht(?!) conjecture is studying the global behavior of a sequence - i.e., not 'does this equation hold in the limit?' but 'does it hold everywhere'. I believe you can show easily that the conjecture is true 'in the limit' (in that the telescoped inequality is true)... – Steven Stadnicki Apr 14 '15 at 3:59
• ...but in some sense the conjecture (much like the Riemann hypothesis) is concerned with showing that no outlier behavior can occur, and asymptotic analysis of this sort can't show that outliers can't occur. – Steven Stadnicki Apr 14 '15 at 4:00
• @StevenStadnicki You said "I believe you can show easily that the conjecture is true 'in the limit' (in that the telescoped inequality is true)". Please feel free to post how you would do it. – John Nicholson Apr 14 '15 at 16:12
• The reason I ask is to show that there is a bound to works and to failure. – John Nicholson Apr 14 '15 at 16:27
• John: sure; in this case it's as simple as telescoping the products on either side of that conjecture from $n=1$ to $n=k-1$. You get $\frac{\log p_k}{\log p_1}\lt k$ - but since $p_k\approx k\log k$, the LHS is on the order of some constant times $\log k$, while the RHS is of size $k$. – Steven Stadnicki Apr 14 '15 at 16:35

Let $g=p_{n+1}-p_n$ and write $p=p_n$ for simplicity. Then you're asking for $\log\log(p+g)-\log\log p$ which is $g/p\log p+O(g^2/p^2\log p).$ This
Of course if you want $\sum_{k=1}^n\left(\frac{1}{f(k)}-\log\log p_k\right)=O(1)$ then $\sum_{k=1}^n\frac{1}{f(k)}\sim\sum_{k=1}^n\log\log p_k\sim n\log\log n.$ In fact $\log\log p_k=\log\log k+\log\log k/\log k+O((\log\log k)^2/\log^2k)$ if I am not mistaken, and so the inverse of $f(k)$ should be in this neighborhood. (The first two terms are relevant, the error term should be irrelevant.)