19
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – qwr Mar 14 at 2:18
  • $\begingroup$ 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. $\endgroup$ – qwr Mar 14 at 3:07
  • $\begingroup$ @qwr Seems legit, it is then the question of who will find the answer with worsest bounds for $\pi$. $\endgroup$ – Masterphile Mar 14 at 3:10
  • $\begingroup$ 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. $\endgroup$ – qwr Mar 14 at 3:11
13
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ $8/25$ is nice for calculation since you can write it as $32/100$ and work with decimals. $\endgroup$ – qwr Mar 14 at 4:20
10
$\begingroup$

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 $\pi-3>10/71$.

$\endgroup$
  • $\begingroup$ You used calculator! :D $\endgroup$ – Masterphile Mar 14 at 2:20
  • $\begingroup$ 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 ${\sin}^{-1}$. $\endgroup$ – Masterphile Mar 14 at 2:49
  • $\begingroup$ Does that count as computer software? Rational fractions would involve inelegantly large numbers. Is there a bound on pi that avoids that? $\endgroup$ – Oscar Lanzi Mar 14 at 2:50
  • $\begingroup$ @Ante Now I have an answer which is "purely by hand". $\endgroup$ – WhatsUp Mar 14 at 3:58
6
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ The estimation $\pi > \frac{25}8$ already suffices. It is in some sense an "optimal choice". $\endgroup$ – WhatsUp Mar 14 at 3:22
5
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.