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 ?


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 $
