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?
 A: 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 $\pi-3>10/71$.
A: For any $3.14\le x\le\pi$,
$$
\begin{align}
\sum_{p\in\mathbb{P}}\frac1{\pi^p}
&\le\overbrace{\ \ \frac1{x-1}\ \ }^{\sum\limits_{p=1}^\infty\!\frac1{x^p}}-\overset{\substack{1\not\in\mathbb{P}\\[4pt]\downarrow}\\[4pt]}{\frac1x}-\overset{\substack{4\not\in\mathbb{P}\\[4pt]\downarrow}\\[4pt]}{\frac1{x^4}}&\overset{\substack{x=\frac{333}{106}\lt\pi\\[4pt]\downarrow}\\[14pt]}{0.138374964}&&\overset{\substack{x=3.14\lt\pi\\[6pt]\downarrow}\\[14pt]}{0.138531556}\\
&\le x-3&0.141509434&&0.140000000\\[9pt]
&\le \pi-\lfloor\pi\rfloor&0.141592654&&0.141592654
\end{align}
$$
A: The sum is $\approx 0.137175$ so we have a little bit of leeway since $\pi - 3 = 0.14159...$ The negative exponents of $\pi$ get small very quickly.
Use $47/15 < \pi$ and we can bound with geometric sum of all negative powers $\ge 5$. 
\begin{align}
\sum_{p \ \text{prime}} \pi^{-p} &< \sum_{p \ \text{prime}} (47/15)^{-p}  \\
&= (47/15)^{-2} + (47/15)^{-3} + \sum_{p \ge 5, \ p \  \text{prime}} (47/15)^{-p} \\
&< 
(47/15)^{-2} + (47/15)^{-3} +  \sum_{n = 5}^\infty (47/15)^{-n} \\
&= (47/15)^{-2} + (47/15)^{-3} + \frac{1}{(47/15-1)(47/15)^4}
\end{align}
I used a calculator here but the numbers are doable and the sum is about 0.1392. 
A: 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!
