Is it possible to justify these approximations about prime numbers?

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

$0leqslant g(a)-dfrac{a^2(a^2-1)}{a^2+a-1}leqslant dfrac{a^7}{a^5+a^4+a^2+1} - dfrac{a^2(a^2-1)}{a^2+a-1}$
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{$0leqslant g(a)-dfrac{a^2(a^2-1)}{a^2+a-1}leqslant dfrac{1}{a^5}$}$
And:
$forall a in [2,+infty[ , text{Round} left( g(a)-dfrac{a^2(a^2-1)}{a^2+a-1}right) = 0 $

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

$$ f(a) = frac{1}{a^2} + frac{1}{a^3} + frac{1}{a^5} + Obigg(frac{1}{a^7}bigg) $$

whereas

$$ frac{2a^2}{(a-1)(2a^3 + 2a - 1)} = frac{1}{a^2} + frac{1}{a^3} + frac{1}{2a^5} + frac{3}{2a^6} + Obigg(frac{1}{a^7}bigg). $$

Hence,

$$ f(a) = frac{2a^2}{(a-1)(2a^3 + 2a - 1)} + Obigg(frac{1}{a^5}bigg) $$

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,

$$ f(a) = frac{a^2}{(a-1)(a^3 + a - 1)} + Obigg(frac{1}{a^6}bigg) $$

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

$$ f(a) = frac{a^7 + a^6 + a^4 + a^2 -a - 1}{a^3(a^6 - 1)} + Obigg(frac{1}{a^{25}}bigg) $$

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

$$ nlog n + nloglog n - n < p_n < nlog n + nloglog n $$

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

$$ frac{1}{a^2} + frac{1}{a^3} + frac{1}{a^5} + sum_{n = 6}^{infty}frac{1}{a^{nlog n + nloglog n }} < sum_{n = 1}^{infty} frac{1}{a^{p_n}} < sum_{n = 1}^{infty}frac{1}{a^{nlog n + nloglog n - n }} $$

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.