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?

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.)
