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Let $$S=\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!},\tag1$$ its numeric value is approximately $S \approx 0.517977853388534047...$${}^{[more\ digits]}$

$S$ can be represented in terms of the generalized hypergeometric function: $$S={_3F_2}\left(\frac12,1,1;\ \frac43,\frac53;\ \frac4{27}\right)\cdot\frac12.\tag2$$


Let $\sigma$ be the closed-form expression constructed from integers and elementary functions as follows: $$\sigma=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag3$$ where $$\alpha=\frac{\sqrt[3]{3\,}}{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{9-\sqrt{69}}+\sqrt[3]{9+\sqrt{69}}\right),\tag4$$ $$\beta=\frac1{4\,\sqrt[3]{2\,}}\left(\sqrt[3]{25+3\,\sqrt{69}}+\sqrt[3]{25-3\,\sqrt{69}}\right)-\frac12,\tag5$$ $$\gamma=\frac1{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{57\,\sqrt{69}-459}-\sqrt[3]{57\,\sqrt{69}+459}\right)+\frac12\tag6$$ are the unique real roots of the following cubic equations: $$8\,\alpha^3-2\,\alpha-1=0,\tag7$$ $$64\,\beta^3+96\,\beta^2+36\,\beta-23=0,\tag8$$ $$8\,\gamma^3-12\,\gamma^2+16\,\gamma+11=0.\tag9$$ Equivalently, $$\sigma = \frac{3\,p}{2}\,\ln\big(p+1\big)-\frac{1}{2}\sqrt{\frac{3-p}{p}}\arccos\Big( \frac{p-6}{6p+2}\Big)\tag{10}$$ where $p$ is the plastic constant or the real root of $$p^3-p-1=0\tag{11}$$


It can be numerically checked that the following inequality holds: $$\Big|S-\sigma\Big|<10^{-10^5},\tag{12}$$ I conjecture that the actual difference is the exact zero, and thus $S$ has an elementary closed form: $$\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}\stackrel?=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag{13}$$ I am asking for you help in proving this conjecture.

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    $\begingroup$ You have enough free parameters going on that I think you should be able to get that much precision purely by random chance. Where did this particular guess for the closed form come from? $\endgroup$ – Hurkyl Oct 6 '13 at 7:03
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    $\begingroup$ Send it to Jonathan Borwein at Newcastle (Australia), he'll sort it out. $\endgroup$ – Gerry Myerson Oct 6 '13 at 11:56
  • $\begingroup$ Computations in Maple show $|S - \sigma| < 10^{-10000}$, so $S = \sigma$ is probably true. $\endgroup$ – GEdgar Oct 6 '13 at 13:26
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    $\begingroup$ @Hurkyl Is it possible that all decimal digits of two expressions are exactly the same, but purely by random chance? $\endgroup$ – Vladimir Reshetnikov Oct 6 '13 at 19:07
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    $\begingroup$ @Vladimir: You can get some amazing coincidences if you give yourself enough room to search for them. Given the form of the candidate answer, I don't think one should even bat an eye at a hundred or so digits correct. I don't know how you found this answer, so I don't actually know if ten thousand correct digits is to be expected at random or a significant result. That's part of why I asked where this answer came from. The bigger reason why I asked, is that your methods for finding this result would likely be a very useful aide to those who would seek to prove it, should it be true. $\endgroup$ – Hurkyl Oct 6 '13 at 19:30
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We show that the sum equals $$ \int_0^1 \frac{2-3x}{1-x^2+x^3} dx. $$ This integral is "elementary", but requires expanding the integrand in partial fractions, which in turn requires all the solutions of the cubic polynomial in the denominator; so if one insists on writing everything in radicals then the answer is bound to be complicated. The "conjecture" is surely correct ($10^5$ digits is more than enough for moral certainty, especially since $\alpha,\beta,\gamma$ are all in the field generated by the real root of $1-x^2+x^3$), though it may be an unpleasant and unrewarding exercise to check that the partial-fraction integration yields an equivalent answer. (One also wonders how one could possibly "conjecture" such an answer without some sense of where to look...)

The key is to write each term $n! (2n)! / (3n+2)!$ in terms of the beta integral $a!b!/(a+b+1)! = B(a+1,b+1) = \int_0^1 x^a (1-x)^b dx$. Here we write $n! (2n)! / (3n+2)! = B(2n+1, n+2) / (n+1)$, and sum over $n$ to get $$ \sum_{n=0}^\infty \frac{n! (2n)!} {(3n+2)!} = \int_0^1 \sum_{n=0}^\infty \frac{(x^2-x^3)^{n+1}}{n+1} \frac{dx}{x^2} = -\int_0^1 \log(1-x^2+x^3) \frac{dx}{x^2}. $$ (We easily justify the interchange of infinite sum and definite integral because all integrands are positive on $0<x<1$.) We can now integrate by parts to remove the logarithm: $$ -\int_0^1 \log(1-x^2+x^3) \frac{dx}{x^2} = \int_0^1 \log(1-x^2+x^3) \phantom. d\left(\frac{1}{x}\right) = \int_0^1 \frac1x d(\log(1-x^2+x^3)), $$ in which the integrand simplifies to $(2-3x)/(1-x^2+x^3)$, QED.

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    $\begingroup$ Noam, what would be a reasonable way of turning "moral certainty" into "proof"? (It is enough to verify that the two expressions coincide up to ... many digits because ...) $\endgroup$ – Andrés E. Caicedo Oct 6 '13 at 22:38
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    $\begingroup$ I don't know of such a result, and if it exists it must be harder than a direct proof. The integral is a sum of terms $A \log B$ with $A,B$ in the splitting field $K$ of $x^3-x^2+1$, as is the conjectural formula (using $\arccos z = -i \log(z + i\sqrt{1-z^2})$), so it must be doable by choosing a basis $c_1=1, c_2, c_3, \ldots, c_6$ of $K/\bf Q$, writing each side of the conjecture in the form $\sum_{j=1}^6 A_j c_j \log B_j$, and checking that the $B_j$ agree. $\endgroup$ – Noam D. Elkies Oct 7 '13 at 0:07
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    $\begingroup$ Not sure this helps much, but according to my computer the closed form in the question is equal to the expression I get when I evaluate this integral, so the closed form is correct. (I mean precise calculations with algebraic numbers, not just numerical approximate calculations.) $\endgroup$ – Kirill Oct 7 '13 at 8:01
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    $\begingroup$ Thanks, Noam! The trick with the beta integral for me is a new approach I learnt from your answer. The integral you gave in the beginning of your answer can be evaluated and indeed agrees with the closed form $(3)$ from my question. $\endgroup$ – Vladimir Reshetnikov Oct 7 '13 at 17:42
  • $\begingroup$ Can you kindly look at this question which may generalize this integral? $\endgroup$ – Tito Piezas III Nov 14 '16 at 14:28
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partial answer

Use the method of LINK

Since $$ \int_0^1 t^n (1-t)^{2n}\;dt = \frac{n!(2n)!}{(3n+1)!} , $$ If we write $$ S(x) = \sum_{n=0}^\infty \frac{n!(2n)!}{(3n+2)!}\;x^{3n+2} $$ then $$ S(x) = \int_0^1 \left(\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}\right)dt $$ and $S = S(1)$.
But the derivative with respect to $x$ of $\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}$ is a geometric series. Its sum is a rational function, which may be integrated (with some work or a CAS). Plug in $x=1$. Result (if I copied right):

$$ S = \int_0^1\frac{F(t)}{18(1-t)^{4/3}t^{2/3}}\;dt $$

where

$$ F(t) = \pi \,\sqrt {3}-6\,\sqrt {3}\arctan \left( 2/\sqrt {3}\cdot \left( 1-t \right) ^{2/3}{t}^{1/3}+1/ \sqrt {3} \right) \\ -6\,\ln \left( ({1-t})^{1/3} {t}^{2/3} -\left( 1-t \right) t \right) +3\,\ln \left( t \left( 1-t \right) ^{2}+ \left( 1-t \right) ^{4/3}{t}^{2/3}+ \left( 1-t \right) ^{2/3}{t}^{1/3} \right) \\ +6\,\ln \left( ({1-t})^{1/3}{t}^{2/3} \right) -3\,\ln \left( \left( 1-t \right) ^{2/3}{t}^{1/3} \right) $$

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    $\begingroup$ The last two terms can be simplified. $\endgroup$ – Pedro Tamaroff Oct 6 '13 at 22:35
  • $\begingroup$ The last 4 terms can be simplified to log of a rational function. That didn't seem to help. First 2 terms is a convergent integral, last 4 terms is also a convergent integral. $\endgroup$ – GEdgar Oct 7 '13 at 13:41
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    $\begingroup$ How does any of his compare to the originally conjectured value of S? $\endgroup$ – DaveUM Oct 11 '13 at 16:42
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Alternatively, one can find a simple and elementary solution in logarithms as, $$\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}= -x_1\ln(1-x_1)-x_2\ln(1-x_2)-x_3\ln(1-x_3) = 0.5179778\dots$$ where the $x_n$ are the three roots of $x^3-x+1=0$ and which is the minpoly of the negated plastic constant. See also this post.

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