Happy $\pi$-day! Is it true that $\sum_{p \;\text{prime} } \frac{1}{{\pi}^p} < \pi -\lfloor \pi \rfloor$?

Today is a $\pi$-day and I made this exercise for that purpose (and not only for that!):

Let: $$\phi = \sum_{p \;\text{prime} } \frac{1}{{\pi}^p}$$ By applying only knowledge of calculus and, more generally (if needed), real analysis of functions of one variable, and without computational software, determine is it true that we have: $$\phi< \pi - \lfloor\pi\rfloor$$ Where $\lfloor\pi\rfloor=3$ is the floor function of $\pi$.

Is this possible to solve with, for example, some of the formulas for infinite product for $\pi$ or Taylor series for ${\sin}^{-1}$, without any numerical estimates?

Or, if estimates are needed, what is the worst one you need to apply to solve this?

Among the various methods I tested, the following seems to be the "simplest", in terms of the arithmetic heights of the rational numbers involved.

First step is to use the estimation $\pi > \frac{25}8$, and the fact that all prime numbers, except $2$, are odd.

This yields:

$$\phi < a^2 + a^3(1 + a^2 + \dotsc) = a^2 + \frac{a^3}{1 - a^2},$$ where $a = \frac8{25}$.

With a bit of calculation, we get the rational number on the right hand side: $\frac{48704}{350625}$.

Second step is to use another estimation $\pi > \frac{157}{50}$. This gives $\pi - \lfloor\pi\rfloor > \frac7{50}$.

A final calculation shows that $\frac{48704}{350625} < \frac7{50}$, hence $\phi < \pi - \lfloor\pi\rfloor$.

The only thing remains is to explain the two estimations.

Since a simple calculation shows $\frac{157}{50} > \frac{25}{8}$, we only need to show that $\pi > \frac{157}{50} = 3.14$.

I claim that the OP already knows this: because that's why it's called THE PIE DAY!

I shall assume the following, proved by Archimedes:

$\pi>3\dfrac{10}{71}$

Then the quoted sum is rendered

$\phi< \sum_{n \in \mathbb P : n=2,3,5,7,... } (\frac{71}{223})^n$

Tlhe primes consist of $2, 3,$ and a subset of $\{n\in\mathbb N:6n\pm 1\}$. So $\phi$ is less than the sum of two terms plus two geometric series:

$\phi<(\frac{71}{223})^2+(\frac{71}{223})^3+ \sum_{n \in \mathbb N} (\frac{71}{223})^{6n-1}+ \sum_{n \in \mathbb N} (\frac{71}{223})^{6n+1}$

Summing the last two summations as geometric series gives

$\phi<(\frac{71}{223})^2+(\frac{71}{223})^3+ \frac{1}{1-(71/223)^2}((\frac{71}{223})^5+(\frac{71}{223})^7)$

When this last comparison value is multiplied by $71$ and put into a calculator the result is between $9$ and $10$, so $\phi<10/71$ whereas Archimedes had rendered $\pi-3>10/71$.