Irrationality Proofs and Measures for log(1+b/a) for positive integers a and b such that a>(b-1/e)^2/4



                              By Shalosh B. Ekhad



Theorem: Let a and b be positive integers such that a>(b-1/e)^2/4, then log(1+b/a) is an irrational number

and an irrationality measure is,

                              1/2                              1/2
       2 a + b - 2 (a (a + b))          2 a + b + 2 (a (a + b))
    ln(--------------------------) - ln(--------------------------)
                    2                                2
                   b                                b
    ---------------------------------------------------------------
                                       1/2
                2 a + b - 2 (a (a + b))
             ln(--------------------------) + 2 ln(b) + 1
                             2
                            b



                  In Maple format the irrationality measure is



(ln((2*a+b-2*(a*(a+b))^(1/2))/b^2)-ln((2*a+b+2*(a*(a+b))^(1/2))/b^2))/(ln((2*a+
b-2*(a*(a+b))^(1/2))/b^2)+2*ln(b)+1)


                         Proof: Consider  the integral



                                  1
                                 /     n        n
                                |     x  (1 - x)
                        E(n) =  |   ---------------- dx
                                |            (n + 1)
                               /    (b x + a)
                                 0



                            Let c be the constant

                                  ln(1 + b/a)
                                  -----------
                                       b



                It is readily seen that E(n) can be written as



                               E(n)=A(n)+B(n)*c



             for some sequences A(n) and B(n) of RATIONAL NUMBERS

             It follows from the Almkvist-Zeilberger algorithm that



E(n), and hence A(n) and B(n), all satisfy the following second-order linear   recurrence with polynomial coefficients



                                                    2
     (n + 1) F(n) - (2 n + 3) (2 a + b) F(n + 1) + b  (n + 2) F(n + 2) = 0





                but of course with different initial conditions.





                   It follows from the Poincarre lemma that



              A(n),B(n)=OMEGA(C1(a,b)^n), E(n)=OMEGA( C2(a,b)^n )



where C1(a,b) is the larger root and C2(a,b) the smaller root of the indicial equation of the constant-coefficient

                 recurrence approximating the above recurrence

                       that indicial polynomial, in N, is



                                                 2  2
                           1 + (-4 a - 2 b) N + b  N



                              and in Maple format



1+(-4*a-2*b)*N+b^2*N^2


        Solving this quadratic, we get that the larger root, C1(a,b) is



                                         2       1/2
                           2 a + b + 2 (a  + a b)
                           -------------------------
                                       2
                                      b



                       and the smaller root, C2(a,b),  is



                                           2       1/2
                            -2 a - b + 2 (a  + a b)
                          - --------------------------
                                         2
                                        b



                   and in Maple format,thery are respectively



(2*a+b+2*(a^2+a*b)^(1/2))/b^2, -(-2*a-b+2*(a^2+a*b)^(1/2))/b^2


We now need the following divisibilty lemma that is left as an exercise to the reader.



Lemma: A1(n)=lcm(1..n)*b^(2*n)*A(n) and B1(n)=lcm(1..n)*b^(2*n)*B(n) are always integers.



                    Let E1(n)=lcm(1..n)*b^(2*n)*E(n) , then



                              E1(n)=A1(n)+B1(n)*c



                    where now A1(n) and B1(n) are INTEGERS.



                      Since lcm(1..n)=OMEGA(e^n), we have



    A1(n),B1(n)=OMEGA((b^2*e*C1(a,b))^n ), E(n)=OMEGA( (b^2*e*C2(a,b))^n )



                                     Hence

              E(n)=OMEGA(1/max(A(n),B(n))^delta ), where delta is



                     -log(b^2*e*C2(a,b))/log(b^2*e*C1(a,b)



                                             2       1/2
                              -2 a - b + 2 (a  + a b)
                         ln(- --------------------------) + 2 ln(b) + 1
                                           2
                                          b
         that equals , - ----------------------------------------------
                                           2       1/2
                             2 a + b + 2 (a  + a b)
                          ln(-------------------------) + 2 ln(b) + 1
                                         2
                                        b



                           and in Maple format it is



-(ln(-(-2*a-b+2*(a^2+a*b)^(1/2))/b^2)+2*ln(b)+1)/(ln((2*a+b+2*(a^2+a*b)^(1/2))/
b^2)+2*ln(b)+1)


          It is readily seen that this is positive when a>(b-1/e)^2/4



and an irrationality measure can be taken to be 1+1/de, that establishes the theorem. QED