-3

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)

Community
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Jaume Oliver Lafont
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    When the degree of the denominator of the fraction is low the computations are slow. – FDP Feb 15 '16 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...) – Jaume Oliver Lafont Feb 15 '16 at 10:37
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    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. – Giovanni Resta Feb 15 '16 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 $6009$ – Jaume Oliver Lafont Feb 15 '16 at 10:55
  • @GiovanniResta Even worse: do you exactly hit 333/106 when summing 6009 terms or is there some excess fraction that should be written down at the RHS of $\pi-\frac{333}{106}$? There should be no such a fraction. – Jaume Oliver Lafont Feb 15 '16 at 11:06
  • What about $\displaystyle \dfrac{1}{113}\sum_{k=0}^{+\infty}\dfrac{4281213600+1425552480k}{(4k+5)(4k+7)(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)(4k+19)}$ ? (sorry wrong fraction) – FDP Feb 15 '16 at 11:10
  • The last series is equal to $\pi-\dfrac{333}{113}$ according to Wolfram Alpha. (wrong denominator i know) – FDP Feb 15 '16 at 11:21
  • No, the one i have put in comment. I know its not related to the question but it's nice however. – FDP Feb 15 '16 at 11:23
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    @JaumeOliverLafont No, I meant that summing up 6009 terms you exceed 333/106. I had to shorten the comment because it was too long. – Giovanni Resta Feb 15 '16 at 12:05
  • there is one with constant numerator for $\frac{22}{7}-\pi$ http://math.stackexchange.com/a/1657416/134791 – Jaume Oliver Lafont Feb 15 '16 at 23:07

2 Answers2

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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 $(4k+1)...(4k+8)$)

FDP
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  • 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? – Jaume Oliver Lafont Feb 15 '16 at 11:42
  • Do you see any hope to get rid of the linear term with $k$? – Jaume Oliver Lafont Feb 15 '16 at 13:25
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    Frankly, no. But it's just a feeling, no proof. – FDP Feb 15 '16 at 13:29
  • There are formulas like these with constant numerator for $\pi-3$, $\log(2)-\frac{2}{3}$, $\frac{7}{10}-\log(2)$ and $\log(3)-1$. For instance, can you obtain Lehmer's variant for $\pi-3$ with your algorithm? (I guess I cannot either by hand withouth knowing the result in advance...) – Jaume Oliver Lafont Feb 15 '16 at 13:31
  • This one: $$\pi-3=\sum_{k=1}^\infty \frac{24}{(4k+1)(4k+2)(4k+4)}$$ – Jaume Oliver Lafont Feb 15 '16 at 13:37
  • What is the Lehmer's variant? numerator is $ak^2+bk+c$? – FDP Feb 15 '16 at 13:41
  • The one two comments above, with numerator only $c$. Note the irregular denominator... – Jaume Oliver Lafont Feb 15 '16 at 13:43
  • $\displaystyle \dfrac{24}{(4k+1)(4k+2)(4k+4)}=\dfrac{24(4k+3)}{(4k+1)(4k+2)(4k+3)(4k+4)}$ no more irregularity. Anyway, series with such term (equivalent here to $\dfrac{A}{k^3}$) are slowly convergent. Not easy to deal with GP PARI. Maybe some acceleration of convergence do exist for such series. The problem is to have a value for such series with alot of decimals (at least 30-50). – FDP Feb 15 '16 at 20:23
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    The problem of the slow convergent can be bypassed, i think, in transforming series into integrals $\displaystyle \int_0^1 f(x)dx$ – FDP Feb 16 '16 at 22:18
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    $A=\displaystyle \sum_{k=0}^{+\infty} \dfrac{1}{(4k+1)(4k+3)(4k+2)(4k+4)}=\int_0^1 \dfrac{(1-x)^3}{6(1-x^4)}dx$ $B=\displaystyle \sum_{k=0}^{+\infty} \dfrac{k}{(4k+1)(4k+3)(4k+2)(4k+4)}=\int_0^1 \dfrac{(1-x)^2(4x-1)}{24(1-x^4)}dx$ Evaluation of these integrals with 100 decimals is very fast with PARI GP lindep([Pi,A,B])=[1,-72,-96] – FDP Feb 16 '16 at 23:57
  • @FDP - re integrals ... See http://math.stackexchange.com/questions/860499/seeking-proof-for-the-formula-relating-pi-with-its-convergents – Alex Feb 17 '16 at 01:27
  • @FDP Very interesting. Note there is a $(1-x)$ factor that may simplify so, equivalently, $$A=\frac{1}{3!}\int_0^1 \frac{(1-x)^2}{(1+x)(1+x^2)}dx$$ and $$B=\frac{1}{4!}\int_0^1 \frac{(1-x)(4x-1)}{(1+x)(1+x^2)}dx$$ – Jaume Oliver Lafont Feb 17 '16 at 04:21
  • But those have only $\pi$ and $log(2)$ in the result. For rational approximations of only one constant, one may use at least one of $$A'=\sum_{k=1}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$$ and $$B'=\sum_{k=2}^\infty \frac{k}{(4k+1)(4k+2)(4k+3)(4k+4)}$$ (Note that the $k$ in $B$ makes the first term zero) – Jaume Oliver Lafont Feb 17 '16 at 04:37
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    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 Feb 17 '16 at 09:10
  • Given the series for $\pi-3$ and $\frac{22}{7}-\pi$ maybe $\sum_{k=n_0}^\infty \frac{k(4k+a)(4k+b)(4k+c)}{(4k+1)(4k+2)..(4k+12)}$ with $0<a,b,c<=12$ and positive integer $k$ has some chance to evaluate to $\pi-\frac{333}{106}$ – Jaume Oliver Lafont Feb 22 '16 at 08:27
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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: $$ \frac{22}{7}-\frac{1}{2^8}<\pi <\frac{22}{7}.\tag{3} $$

Jack D'Aurizio
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  • Isn't that much like http://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. – Jaume Oliver Lafont Feb 15 '16 at 11:24