For the original Zeta(2) Sketch of an Irrationality Proof of the constant 3F2([1, 1, 1][2, 2];1) By Shalosh B. Ekhad Theorem: The constant of the title 3F2([1, 1, 1][2, 2];1) let's call it c, that is easily seen to be given by the following Beukers-ty\ pe integral 1 1 / / | | 1 c = | | -------- dx dy | | -x y + 1 / / 0 0 1/2 5 5 ln(11/2 + ------) + 2 2 is irrational, with an irrationality measure, 1 + --------------------- 1/2 5 5 ln(11/2 + ------) - 2 2 that in decimals equals, 11.85078220 2 Pi Comment: Note that this constant appears to be , --- 6 Prove it! We need two lemmas Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\ fy the second-order recurrence 2 2 (1 + n) X(n) (11 n + 33 n + 25) X(1 + n) - ------------- + ---------------------------- + X(2 + n) = 0 2 2 (2 + n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -5 B(0) = 1, B(1) = -3 A(n) Then, ----, approximates the constant of the title, 3F2([1, 1, 1][2, 2];1) B(n) let's call it c, that can be easily seen to be given by the Beukers-type in\ tegral 1 1 / / | | 1 c = | | -------- dx dy | | -x y + 1 / / 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof, consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / F(n) = | | ---------------------- dx dy | | -x y + 1 / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma TWO: Let E(n) be 2 LCM(n) where LCM(1..m) is the least-common-multiple of the first m integers, and Then there exists a sequence G(n) of integers that is asymptotically negligi\ ble, i.e. the limit of log( G(n) )/n tends to 0 as n goes to infinity such that A1(n):=E(n)G(n)A(n), B1(n):=E(n)G(n)B(n) are BOTH integers For example, G(n) for n=, 3000, equals 1 whose log divided by n is 0. Indeed peanuts. Using standard asymptotics derived from the Prime Number Theorem it is readi\ ly seen that the limit of log(E(n))/n , and hence log(E(n)*G(n))/n, equals exactly, 2 that equals in decimals, 2. Just as a sanity check, the empircal numerical values of nu for E(n) from, 2980, to , 3001, are [1.998222771, 2.000030962, 2.002963467, 1.998725255, 1.988565113, 1.997098998, 1.998674236, 1.996311598, 1.993856404, 1.997233516, 1.994378823, 1.995007588, 1.993000121, 1.994544935, 1.995669433, 1.994636284, 1.992117492, 1.991704999, 1.982433949, 1.993890461, 1.991231289] Multiplying F(n) by E(n)*G(n) we get E(n) F(n) G(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - 2 2 where , delta = --------------------- 1/2 5 5 ln(11/2 + ------) + 2 2 that in decimals is, , 0.09215925467 As you can see, delta is positive We leave it to the reader to fill-in the details. -------------------------------------------------- ----------------------- The following examples give positive delta, i.e. yielding irrationality proo\ fs For GBC(0,0,1/2,0) (alias 2*log(2)) Sketch of an Irrationality Proof of the constant 3F2([1, 1, 1/2][2, 3/2];1) By Shalosh B. Ekhad Theorem: The constant of the title 3F2([1, 1, 1/2][2, 3/2];1) let's call it c, that is easily seen to be given by the following Beukers-ty\ pe integral 1 1 / / | | 1 c = 1/2 | | --------------- dx dy | | 1/2 / / y (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + 2 2 is irrational, with an irrationality measure, 1 + --------------------- 1/2 5 5 ln(11/2 + ------) - 2 2 that in decimals equals, 11.85078220 Comment: Note that this constant appears to be , arcsinh(15/8) Prove it! We need two lemmas Lemma ONE: , let A(n), B(n), be two sequences of rational numbers that satis\ fy the second-order recurrence 2 4 (1 + n) (1 + 2 n) (10 n + 17) X(n) - ------------------------------------- 2 (10 n + 7) (2 n + 5) (3 + 2 n) 3 2 2 (220 n + 814 n + 950 n + 341) X(1 + n) + ------------------------------------------ + X(2 + n) = 0 (10 n + 7) (2 n + 5) (3 + 2 n) Subject to the initial conditions A(0) = 0, A(1) = -14/3 B(0) = 1, B(1) = -10/3 A(n) Then, ----, approximates the constant of the title, 3F2([1, 1, 1/2][2, 3/2];1) B(n) let's call it c, that can be easily seen to be given by the Beukers-type in\ tegral 1 1 / / | | 1 c = 1/2 | | --------------- dx dy | | 1/2 / / y (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof, consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / F(n) = 1/2 | | ---------------------- dx dy | | 1/2 | | y (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma TWO: Let E(n) be 2 PP(0, 1/2, 0, 1, {0}, 1, n) PP(1/2, 1, 0, 2, {0}, 1, n) -------------------------------------------------------- (2 n) 2 where for 0