# A series of positive terms to prove $\pi>\frac{333}{106}$

This is a consequence of the answer to that question.

A proof that $\pi > \frac{333}{106}$ is given by the series of positive terms $$\pi-\frac{333}{106} \\ =\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)}$$

Q Is there a similar series for $\pi-\frac{333}{106}$ with a lower degree in the numerator?

(The accepted answer shall provide either a proof or PARI code)

• When the degree of the denominator of the fraction is low the computations are slow.
– FDP
Commented Feb 15, 2016 at 10:08
• I got stuck sometimes before I understood that initial shift mod 4 has to be 1 for convergents from below (in (4k+9), 9 mod 4=1) and it has to be 3 for convergents from above (for 22/7, 355/113, etc...) Commented Feb 15, 2016 at 10:37
• Did you mean "numerator" or "denominator"? In any case, I don't see the problem. You take a known summation for $\pi$ with positive terms, for example $\sum_{k\ge 1}8/((4 k - 3) (4 k - 1))$. This has numerator of degree 0 and denom. of degree 2. Then you add enough terms to reach $333/106$ (in this case you need 6009 terms). The rest of the summation provides a proof that $\pi > 333/106$. And if you really want you can shift the formula to have the index starting from 0 or 1. Unfortunately, often the smaller the degrees of the polynomials involved, the larger the number of terms needed. Commented Feb 15, 2016 at 10:49
• @GiovanniResta I meant numerator. The terms I would like to see out are $118720k^2+762311k$ The goal is writing a series as elegant as the one for $\pi-3$ $$\pi-3=\sum_{k=1}^\infty \frac{24}{(4 k+1) (4 k+2) (4 k+4)}$$ with no large numbers such as $6009$ Commented Feb 15, 2016 at 10:55
• @JaumeOliverLafont No, I meant that summing up 6009 terms you exceed 333/106. I had to shorten the comment because it was too long. Commented Feb 15, 2016 at 12:05

According to Wolfram Alpha:

$\displaystyle \dfrac{1}{371}\sum_{k=0}^{+\infty}\dfrac{74704320+14936400k}{(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)(4k+19)(4k+21)(4k+23)}=\pi-\dfrac{333}{106}$

Another equality, i post it because it's nice and the "shift" of the denominator is from $7$ to $21$:

$\displaystyle \dfrac{1}{53}\sum_{k=0}^{+\infty}\dfrac{15864560k+4208000k^2}{(4k+7)(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)(4k+19)(4k+21)}=\pi-\dfrac{333}{106}$

This one is nice too:

$\displaystyle \dfrac{1}{53}\sum_{k=0}^{+\infty}\dfrac{3313800+1557360k}{(4k+7)(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)(4k+19)(4k+21)}=\pi-\dfrac{333}{106}$

Some explanations:

Consider:

$\displaystyle R(m,n)=\sum_{k=0}^{\infty}\dfrac{1}{\prod_{r=m}^{n}(4k+2r+1)}$ $\displaystyle S(m,n)=\sum_{k=0}^{\infty}\dfrac{k}{\prod_{r=m}^{n}(4k+2r+1)}$ $\displaystyle T(m,n)=\sum_{k=0}^{\infty}\dfrac{k^2}{\prod_{r=m}^{n}(4k+2r+1)}$

One search for such linear integral relation: $a(\pi-\dfrac{333}{106})+b\times R(m,n)+c\times S(m,n)+d\times T(m,n)=0$

$a,b,c,d$ are integers not necessary positive.

Sometimes (always?), there exist integers $b,c,d$ such that: $b\times R(m,n)+c\times S(m,n)+d\times T(m,n)=0$

To get positive coefficients (when it's possible) search for relations between $\pi-\dfrac{333}{106}$ and two of the $R,S,T$.

PARI GP commands:

suminf(k=0,...) to compute series.

prod(k=m,n,...) to compute products.

\p 100 (changing precision to 100 decimals for example)

lindep([r,s,t]) to find linear integral relation. (search for $a,b,c$ such that $ar+bs+ct=0$)

For example, the last series have been obtained using the command:

lindep([Pi-333/106,R(3,10),S(3,10)])

Sometimes you get false solution due to precision.

Increase the precision and launch again lindep to see if the coefficients are still the same.

Beware the computations can be slow (low "shift" especially and the use of $U(m,n)$). All my computations have been made with a "shift" of $8$. (for example $(4k+1)...(4k+8)$)

• I would love to know how to build your simpler results. Could you show how to construct it (example) or provide some PARI code, please? Commented Feb 15, 2016 at 11:42
• Frankly, no. But it's just a feeling, no proof.
– FDP
Commented Feb 15, 2016 at 13:29
• The problem of the slow convergent can be bypassed, i think, in transforming series into integrals $\displaystyle \int_0^1 f(x)dx$
– FDP
Commented Feb 16, 2016 at 22:18
• $A=\displaystyle \sum_{k=0}^{+\infty} \dfrac{1}{(4k+1)(4k+3)(4k+2)(4k+4)}=\int_0^1 \dfrac{(1-x)^3}{6(1-x^4)}dx$ $B=\displaystyle \sum_{k=0}^{+\infty} \dfrac{k}{(4k+1)(4k+3)(4k+2)(4k+4)}=\int_0^1 \dfrac{(1-x)^2(4x-1)}{24(1-x^4)}dx$ Evaluation of these integrals with 100 decimals is very fast with PARI GP lindep([Pi,A,B])=[1,-72,-96]
– FDP
Commented Feb 16, 2016 at 23:57
• What is weird for me, is that the series with numerator 1,k,k^2 are linearly integral dependant. Series $A=\dfrac{1}{24}(6\log(2)-\pi)$ and $B=\dfrac{1}{48}(2\pi-9\log(2))$, WA computes them easily thus, no need LLL stuff, solving a simple linear system 2x2 is enough.
– FDP
Commented Feb 17, 2016 at 9:10

This is not a series approach, but an integral approach. We may check that: $$\int_{0}^{1}\frac{x^8 (1-x)^8}{1+x^2}\,dx = 4\pi-\frac{188684}{15015} \tag{1}$$ but the integrand function is a non-negative function on $(0,1)$, bounded by $\frac{1}{2^{16}}$. It follows that: $$\pi > \frac{47171}{15015} > \frac{333}{106}.\tag{2}$$ By replacing the exponent $8$ in the LHS of $(1)$ with $4$, we recover the Archimedean approximation: $$\frac{22}{7}-\frac{1}{2^8}<\pi <\frac{22}{7}.\tag{3}$$

• Isn't that much like math.stackexchange.com/a/1593090/134791? I didn't bound the error, though... Here the attempt is to exactly match the difference, with no excess. Commented Feb 15, 2016 at 11:24