10
$\begingroup$

This extends this post.

I. For $\pi^3$:

$$\pi^6-31^2 =\sum_{k=0}^\infty\left(-\frac{63}{(2k+2)^6}+\frac{31^2}{(2k+3)^6}\right) =\sum_{k=0}^\infty P_1(k)\tag1$$

As pointed out by J. Lafont, when $P_1(k)$ is expanded out, its coefficients are all positive. Thus so is the $\text{LHS}$, implying $\pi^3>31$.

II. For $\pi^4$:

The convergents of $\pi^4$ are,

$$97,\, \frac{195}{2},\, \frac{487}{5},\, \frac{1656}{17},\, \frac{2143}{22},\dots$$

The last one, being the particularly close approximation $22\pi^4 \approx 2143.0000027$, was mentioned by Ramanujan. (See also this post.) Using,

$$\frac{\pi^8}{9450}=\sum_{k=0}^\infty \frac{1}{(k+1)^8}$$

$$\frac{17\pi^8}{161280}=\sum_{k=0}^\infty \frac{1}{(2k+1)^8}$$

and the same method to find $(1)$, we get,

$$\pi^8-\Big(\frac{487}{5}\Big)^2 =\sum_{k=0}^\infty\left(\frac{381}{5(2k+2)^8}+\frac{r_1^2}{(2k+3)^8}\right)=\sum_{k=0}^\infty P_2(k)\tag2$$

$$\pi^8-\Big(\frac{2143}{22}\Big)^2 =\sum_{k=0}^\infty\left(-\frac{181695}{11^2(2k+2)^8}+\frac{r_2^2}{(2k+3)^8}\right)=\sum_{k=0}^\infty Q_1(k)\tag3$$

$$\pi^8-\Big(\frac{2143}{22}\Big)^2 =\sum_{k=0}^\infty\left(\frac{r_2^2}{(k+2)^8}-\frac{70208}{1815(2k+1)^8} \right)=\sum_{k=0}^\infty Q_2(k)\tag4$$

where $r_1 =\frac{487}{5},\,$ $r_2 =\frac{2143}{22}$. The coefficients of $P_2(k)$ are all positive, so $5\pi^4>487$.

However, when the $Q_i(k)$ are expanded out, the constant term for both is negative, so we cannot make an analogous conclusion. (In fact, it takes several terms before the sum turns positive.)

Q: Can one find a similar series for $\pi^8-\Big(\frac{2143}{22}\Big)^2 = \sum_{k=0}^\infty R(k)$ such that all coefficients are positive and immediately implying $22\pi^4>2143$?

$\endgroup$
10
  • $\begingroup$ Like the phrase "extends this post" first time seen this, nice $\endgroup$
    – jimjim
    Feb 11, 2016 at 0:41
  • $\begingroup$ @AkivaWeinberger:Oops, thanks. $\endgroup$ Feb 11, 2016 at 1:53
  • 1
    $\begingroup$ If nothing comes out for $\pi^4-\frac{2143}{22}$ from$\pi^8$, how about $\pi^{12}$? $\endgroup$ Feb 11, 2016 at 17:04
  • 1
    $\begingroup$ Another solution might come from adding a suitable zero relation, similarly to changing Gregory-Leibniz series for $\frac{\pi}{4}$ into Lehmer's for $\frac{\pi}{3}$, which proves $\pi>3$ with one term. math.stackexchange.com/questions/14113/… $\endgroup$ Feb 11, 2016 at 17:13
  • 1
    $\begingroup$ @JaumeOliverLafont: You're right, $\pi^{12}$ can be used to prove $\pi^4 >\frac{2143}{22}$. Thanks. :) $\endgroup$ Feb 11, 2016 at 23:32

2 Answers 2

4
$\begingroup$

Thanks to a clever suggestion by J. Lafont, there is a series that can prove $\displaystyle\pi^4>\frac{2143}{22}$. However, it does not use $\pi^8$ but $\pi^{12}$. We start with,

$$\frac{691\pi^{12}}{638512875} = \sum_{k=0}^\infty \frac{1}{(k+1)^{12}}$$

$$\frac{691\pi^{12}}{638668800}-1 = \sum_{k=0}^\infty \frac{1}{(2k+3)^{12}}$$

Multiply them with unknowns $a,b,$ then add the two,

$$\frac{691}{420}\frac{(4096a+4095b)\pi^{12}}{13!}-b = \sum_{k=0}^\infty \left(\frac{a}{(k+1)^{12}}+\frac{b}{(2k+3)^{12}}\right)$$

Let $b=\big(\frac{2143}{22}\big)^3$, and choose $a$ such that $\pi^{12}$ has a unit coefficient. We then get,

$$\begin{aligned}\pi^{12}-\Big(\frac{2143}{22}\Big)^3 &=\sum_{k=0}^\infty\left( -\frac{52410418515}{691\cdot10648\cdot(2k+2)^{12}} +\Big(\frac{2143}{22}\Big)^3\frac{1}{(2k+3)^{12}}\right)\\ &=\sum_{k=0}^\infty R(k) \end{aligned}$$

When expanded out, the coefficients of $R(k)$ are all positive. Thus, the $\text{LHS}$ must be positive. Since it is a difference of two cubes $p^3-q^3 = (p-q)(p^2+pq+q^2)$, then that implies $\displaystyle\pi^4>\frac{2143}{22}.$

$\endgroup$
2
  • $\begingroup$ I revised the series for $\pi^6-961$, so now there is a version with positive coefficients that does not need any further expansion. Would you like to try that improved method on $\pi^{12}-\left(\frac{2143}{22}\right)^3$? Here it is: math.stackexchange.com/a/1651175/134791 $\endgroup$ Feb 15, 2016 at 18:14
  • $\begingroup$ @JaumeOliverLafont: I saw the edited version. Ah, quite nice. $\endgroup$ Feb 15, 2016 at 18:28
3
$\begingroup$

From the accelerated series $$\zeta(4)=\frac{\pi^4}{90}=\frac{36}{17}\sum_{n=1}^{\infty } \frac{1}{n^{4}\dbinom{2n}{n}}$$ (Convergence acceleration technique for $\zeta(4)$ (or for $\eta(4)$) via creative telescoping?)

we have the direct sum

$$\pi^4 -\frac{2143}{22}= \frac{5}{52898832} \sum_{n=10}^\infty \left( \frac{1998926767}{n^4\dbinom{2n}{n}}+\frac{17452241}{(n+1)^4\dbinom{2(n+1)}{n+1}}\right)$$

The denominator in the coefficient fraction factors into small primes: $$52898832=(2·3·7)^4·17$$

This series may be related to these.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .