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?

• use letter $p$ instead of $n$. also since you explained what set P is you don't need to write out examples. just some tips to make the notation cleaner – qwr Mar 14 at 2:18
• there might be a way to use more complicated formulae for pi, but at the end your desired result is a numerical bound so you will probably eventually have to use a calculator somewhere. – qwr Mar 14 at 3:07
• @qwr Seems legit, it is then the question of who will find the answer with worsest bounds for $\pi$. – user757601 Mar 14 at 3:10
• you can use less precise bounds for pi if you are willing to calculate more terms or use some more cleverness (like Oscar Lanzi's answer where he used geometric series $6n \pm 1$). I give a simple answer. – qwr Mar 14 at 3:11

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!

• $8/25$ is nice for calculation since you can write it as $32/100$ and work with decimals. – qwr Mar 14 at 4:20

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 $$\pi-3>10/71$$.

• I do not know, I am trying to solve this without numerical estimates, with infinite product for $\pi$, or at least with Taylor series for ${\sin}^{-1}$. – user757601 Mar 14 at 2:49
• Does that count as computer software? Rational fractions would involve inelegantly large numbers. Is there a bound on pi that avoids that? – Oscar Lanzi Mar 14 at 2:50
• @Ante Now I have an answer which is "purely by hand". – WhatsUp Mar 14 at 3:58

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.

• The estimation $\pi > \frac{25}8$ already suffices. It is in some sense an "optimal choice". – WhatsUp Mar 14 at 3:22

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.

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}