TITLE: Happy $\pi$-day! Is it true that $\sum_{p \;\text{prime} } \frac{1}{{\pi}^p} < \pi -\lfloor \pi \rfloor$?
QUESTION [21 upvotes]: 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?

REPLY [6 votes]: For any $3.14\le x\le\pi$,
$$
\begin{align}
\sum_{p\in\mathbb{P}}\frac1{\pi^p}
&\le\overbrace{\ \ \frac1{x-1}\ \ }^{\sum\limits_{p=1}^\infty\!\frac1{x^p}}-\overset{\substack{1\not\in\mathbb{P}\\[4pt]\downarrow}\\[4pt]}{\frac1x}-\overset{\substack{4\not\in\mathbb{P}\\[4pt]\downarrow}\\[4pt]}{\frac1{x^4}}&\overset{\substack{x=\frac{333}{106}\lt\pi\\[4pt]\downarrow}\\[14pt]}{0.138374964}&&\overset{\substack{x=3.14\lt\pi\\[6pt]\downarrow}\\[14pt]}{0.138531556}\\
&\le x-3&0.141509434&&0.140000000\\[9pt]
&\le \pi-\lfloor\pi\rfloor&0.141592654&&0.141592654
\end{align}
$$