Is it possible to justify these approximations about prime numbers? A recently closed question asked for a possible closed form of the infinite summation
$$f(a)=\sum _{i=1}^{\infty } a^{-p_i}$$ for which I already proposed a first simple but totally empirical approximation.
Since we quickly face very small numbers, I tried to find approximations of
$$g(a)=\Big[\sum _{i=1}^{\infty } a^{-p_i}\Big]^{-1} \qquad \text{and} \qquad h(a)=\Big[\sum _{i=1}^{\infty } (-1)^{i-1} a^{-p_i}\Big]^{-1}$$
All calculations where done with integer values of $a$ for the range $2 \leq a \leq 1000$.
What I obtained is
$$\color{blue}{g(a)\sim\frac{(a-1) (2a^3+2a-1)}{2 a^2}}\qquad \text{and} \qquad \color{blue}{h(a)\sim\frac{(a-1) \left(a^3+2 a^2+3 a+4\right)}{a^2}}$$
If the corresponding curve fits were done, in both cases we should have $R^2 > 0.999999999$.
For the investigated values of $a$,
$$\text{Round}\left[\frac{(a-1) (2a^3+2a-1)}{2 a^2}-{g(a)}\right]=0$$
$$\text{Round}\left[\frac{(a-1) \left(a^3+2 a^2+3 a+4\right)}{a^2}-{h(a)}\right]=0$$

Not being very used to work with prime numbers, is there any way to justify, even partly, these approximations ?

 A: I found this estimate for $g$:  $\ g(a)\sim \dfrac{a^2(a^2-1)}{a^2+a-1}$.
First inequality
$f(a) = \displaystyle \sum_{i=1}^{+\infty} \dfrac{1}{a^{p_i}} \leqslant \sum_{i=2}^{+\infty} \dfrac{1}{a^i} -\sum_{i=2}^{+\infty} \dfrac{1}{a^{2i}} = \dfrac{1}{a^2}\dfrac{1}{1-\dfrac{1}{a}} -\dfrac{1}{a^4}\dfrac{1}{1-\dfrac{1}{a^2}}$
$f(a)\leqslant \dfrac{1}{a(a-1)}-\dfrac{1}{a^2(a^2-1)}= \dfrac{a^2+a-1}{a^2(a^2-1)}$
$\fbox{$g(a)\geqslant \dfrac{a^2(a^2-1)}{a^2-a+1}$}$
Second inequality
$f(a) \geqslant \dfrac{1}{a^2}+\dfrac{1}{a^3}+\dfrac{1}{a^5}+\dfrac{1}{a^7}$
$\fbox{$g(a)\leqslant \dfrac{a^7}{a^5+a^4+a^2+1}$}$
Quality of the approximation
$0\leqslant g(a)-\dfrac{a^2(a^2-1)}{a^2+a-1}\leqslant \dfrac{a^7}{a^5+a^4+a^2+1} - \dfrac{a^2(a^2-1)}{a^2+a-1}$
And:
$\dfrac{a^7}{a^5+a^4+a^2+1} - \dfrac{a^2(a^2-1)}{a^2+a-1} = \dfrac{a^2}{(a^5+a^4+a^2+1)(a^2+a-1)}$
So
$\fbox{$0\leqslant g(a)-\dfrac{a^2(a^2-1)}{a^2+a-1}\leqslant \dfrac{1}{a^5}$}$
And:
$\forall a \in [2,+\infty[ \ , \ \text{Round} \left( g(a)-\dfrac{a^2(a^2-1)}{a^2+a-1}\right) = 0 $
A: These estimates are correct within a reasonable degree of accuracy. Below is the explanation for $f(a)$; the case for $h(a)$ can be dealt similarly. We have
$$
f(a) = \frac{1}{a^2} + \frac{1}{a^3} + \frac{1}{a^5} + O\bigg(\frac{1}{a^7}\bigg)
$$
whereas
$$
\frac{2a^2}{(a-1)(2a^3 + 2a - 1)} = \frac{1}{a^2} + \frac{1}{a^3} + \frac{1}{2a^5} + \frac{3}{2a^6} + O\bigg(\frac{1}{a^7}\bigg).
$$
Hence,
$$
f(a) = \frac{2a^2}{(a-1)(2a^3 + 2a - 1)} + O\bigg(\frac{1}{a^5}\bigg)
$$
For large values of $a$ the error would obviously be negligible, since it grows no faster than a constant times $a^{-5}$. So this may or may not be a good estimate depending upon weather you are satisfied with the magnitude of the error term $O(a^{-5})$.
The best possible estimate of the form $\dfrac{Ax^2}{(x-1)(Bx^3 + Cx^2 + Dx + E)}$ is obtained by the Laurent series expansion of about the point $x = \infty$ and equating the coefficient of smallest non prime powers to zero which gives $A = B = D = 1, C = 0,E = -1$.
Hence we have,
$$
f(a) = \frac{a^2}{(a-1)(a^3 + a - 1)} + O\bigg(\frac{1}{a^6}\bigg)
$$
which reduces the error by a factor of $a$.
Update 21-Jul-2020: However, using basic properties of primes we can get remarkably sharper estimates. Since every primes $\ge 5$ are of the form $6k \pm 1$, by summing up the geometric sequences $a^{-6k-1} + a^{-6k+1}$ for $k = 1,2,\ldots, \infty$ and adding $a^{-2} + a^{-3}$, and taking advantage of the fact the the density of primes among the first few numbers of these form is high we get
$$
f(a) = \frac{a^7 + a^6 + a^4 + a^2 -a - 1}{a^3(a^6 - 1)} + O\bigg(\frac{1}{a^{25}}\bigg)
$$
A: This a long comment. Here is a possible approach to estimate $f(a)$. Using Dusart's approximation (later improved by Axler), The $n$-th prime satisfies
$$
n\log n  + n\log\log n - n < p_n < n\log n  + n\log\log n 
$$
where the lower bound holds for all $n \ge 1$ and the upper bound holds for $n \ge 6$. Hence for $a > 1$, we obtain an inequality of the form
$$
\frac{1}{a^2} + \frac{1}{a^3} + \frac{1}{a^5} + \sum_{n = 6}^{\infty}\frac{1}{a^{n\log n  + n\log\log n }} 
< \sum_{n = 1}^{\infty} \frac{1}{a^{p_n}} 
< \sum_{n = 1}^{\infty}\frac{1}{a^{n\log n  + n\log\log n - n }}
$$
This can gives some tight approximations if we can convert left and the right sums to a closed form approximation with controllable error terms, which however is the more tedious task.
