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

Question

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?

Answer 1

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 $\pi-3>10/71$.

Answer 2

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.

Answer 3

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!

Answer 4

I've spent far too long to not post this. This proof uses only the fact that $\pi>3.14$ and standard calculus.

Initially, $$\phi\le \frac{1}{\pi^2}+\frac{1}{\pi^3}+\frac{1}{\pi^4}+\cdots=\frac{1/\pi^2}{1-1/\pi}=\frac{1}{\pi^2-\pi} \quad(\approx 0.1486)$$ which is quite close. Now we have over-counted the terms $$\frac{1}{\pi^4}+\frac{1}{\pi^6}+\frac{1}{\pi^8}+\cdots=\frac{1/\pi^4}{1-1/\pi^2}=\frac{1}{\pi^4-\pi^2}\quad (\approx 0.0114)$$

It remains to see without a calculator that $$\frac{1}{\pi^2-\pi}-\frac{1}{\pi^4-\pi^2}<\pi-3.$$

Routine manipulation shows this is equivalent to $$\pi^5-3\pi^4-\pi^3+2\pi^2-\pi+1>0.$$

To this end, define $f(x)=x^5-3x^4-x^3+2x^2-x+1$. It suffices to prove the following:

1. $f(3.14)>0$
2. $f$ is increasing on $x>3$.
3. $\pi>3.14$

Proof of $1$: let $k=7/50$, so that $3.14=3+k$. Then

\begin{align} f(3.14)&=(3+k)^5-3(3+k)^4-(3+k)^3+2(3+k)^2-3-k+1\\ &=(3+k)^5-(3+k)(3+k)^4+k(3+k)^4-(3+k)^3+2(3+k)^2-3-k+1\\ &=k(3+k)^4-(3+k)^3+2(3+k)^2-3-k+1\\ &=k\left(3^4+4\cdot 3^3k+6\cdot 3^2k^2+4\cdot 3k^3+k^4\right)\\ &\quad -\left(3^3+3\cdot 3^2k+3\cdot 3k^2+k^3 \right)\\ &\quad +2(9+6k+k^2)\\ &\quad -k-2\\ &=k^5+12k^4+k^3(6\cdot 9-1)+k^2(4\cdot 27-9+2)+k(81-27+12-1)-27+18-2\\ &=k^5+12k^4+53k^3+101k^2+65k-11. \end{align}

Next, \begin{align} k^5+12k^4+53k^3+101k^2+65k-11&>100k^2+65k-11\\ &=\frac{2\cdot 7^2}{50}+\frac{65\cdot 7}{50}-\frac{11\cdot 50}{50}\\ &=\frac{98+420+35-550}{50}\\ &=\frac{3}{50}\\ &>0. \end{align}

Proof of $2$: we have $f'(x)=5x^4-12x^3-3x^2+4x$, so $$f'(3)=5\cdot3^4-4\cdot 3^4-3^3+12=81-27+12>0.$$ Also, $f''(x)=20x^3-36x^2-6x+4$, so $$f''(3)=20\cdot 3^3-12\cdot 3^3-18+4=8\cdot 27-18+4>0.$$ Finally, $f'''(x)=60x^2-72x-6$, so $$f'''(3)=60\cdot 2^2-24\cdot 3^2-6>0.$$ Clearly $f''''(x)=120x-72$ is positive for $x>3$, whence $f'''$ is positive and increasing on $(3,\infty)$. Similarly, $f''$ is positive and increasing for $x>3$, as is $f'$, and thus $f$ is increasing on $(3,\infty)$.

Proof of $3$: exercise.

Answer 5

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