From: Robin Chapman Subject: Re: Harmonic number & central binomial coefficient sum Date: Tue, 04 May 1999 15:06:28 GMT Newsgroups: sci.math To: qqquet@hotbot.com Keywords: Application of Zeilberger's algorithm In article <7gm8s4$lpn$1@nnrp1.dejanews.com>, Robin Chapman wrote: > In article , > qqquet@hotbot.com (Leroy Quet) wrote: > > Here is a closed expression for a sum that I found (not completely on > > my own). > > Let -1/4 > Let B(2m,m)=(2m)!/m!^2. > > Let H(m)=1+1/2+1/3+...1/m. > > Then: > > sum(m=1 to infinity)[B(2m,m)*H(m)*y^m]= > > ln(4y^2/((1-4y)(1-sqrt(1-4y))^2))/sqrt(1-4y) > > I think you mean sum_{m=1}^infinity B(2m,m) H(2m) y^m. One can prove this by using Zeilberger's algorithm. The generating function equals f(y) = [log((1+sqrt(1-4y))/2) - log(1-4y)]/sqrt(1-4y). If g(y) = log((1 + sqrt(1-4y))/2) then f'(y) = [1 - 1/sqrt(1-4y)]/(2y) so the y^n coefficient of the series for f(y) is - B(2m,m)/(2m). Thus the y^n coefficient of f(y) is sum_{k=1}^n (1/k)[4^k - B(2k,k)/2] B(2n-2k, n-k). We sum this by using Zeilberger's algorithm. I split this into sum_{k=1}^n (1/k) 4^k B(2n-2k, n-k) and sum_{k=1}^n (1/k) B(2k,k) B(2n-2k, n-k) and used Zeilberger's MAPLE package EKHAD on both expressions (http://www.math.temple.edu/~zeilberg/programs.html). Putting the results together gives for a(n, k) = (1/k)[4^k - B(2k,k)/2] B(2n-2k, n-k) and b(n, k) = [4^k - (k/2(n+1)) B(2k,k)] B(2n-2k+2, n-k+1) the equation b(n, k+1) - b(n,k) = (4n+2) a(n, k) - (n+1) a(n+1, k). (*) If we set S(n) = sum_{k=1}^n a(n,k) then adding (*) from k = 1 to n gives, after some manipulation, (n+1)S(n+1) - (4n+2)S(n) = [4 - 1/(n+1)] B(2n,n) and by induction then we get S(n) = H_{2n} B(2n,n). Robin Chapman + "They did not have proper School of Mathematical Sciences - palms at home in Exeter." University of Exeter, EX4 4QE, UK + rjc@maths.exeter.ac.uk - Peter Carey, http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda, chapter 20 -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own ============================================================================== From: Robin Chapman Subject: Re: Harmonic number & central binomial coefficient sum Date: Tue, 04 May 1999 17:39:07 GMT Newsgroups: sci.math To: qqquet@hotbot.com In article , qqquet@hotbot.com (Leroy Quet) wrote: > Here is a closed expression for a sum that I found (not completely on > my own). > Let -1/4 Let B(2m,m)=(2m)!/m!^2. > Let H(m)=1+1/2+1/3+...1/m. > Then: > sum(m=1 to infinity)[B(2m,m)*H(m)*y^m]= > ln(4y^2/((1-4y)(1-sqrt(1-4y))^2))/sqrt(1-4y) This function equals [2 log((1 + sqrt(1-4y))/2) - log(1-4y)]/sqrt(1-4y). I've already posted a solution, but here's a simpler one. If f(y) = a_0 + a_1 y + a_2 y^2 + ... then (n+1) a_{n+1} - (4n+2) a_n is the coefficient of y^n in (1-4y) f'(y) - 2 f(y) = sqrt(1-4y) (d/dy) (f(y) sqrt(1-4y)). First let's take f(y) = log((1 + sqrt(1-4y))/2)/sqrt(1-4y). Then sqrt(1-4y) (d/dy) (f(y) sqrt(1-4y)) = -(1 - sqrt(1-4y))/2y and so (n+1) a_{n+1} - (4n+2) a_n = - B(2n, n)/(n+1). Now set a_n = b_n B(2n, n). Then (4n+2)(b_{n+1} - b_n) = -1/(n+1) or b_{n+1} - b_n = -1/(2n+1) + 1/(2n+2). It follows that b_n = -(1 - 1/2 + 1/3 - 1/4 + ... - 1/2n) = -(H_{2n} - H_n) and so a_n = -(H_{2n} - H_n) B(2n, n). Now take f(y) = log(1-4y)/sqrt(1-4y). This time sqrt(1-4y) (d/dy) (f(y) sqrt(1-4y)) = -4/sqrt(1-4y) so that (n+1) a_{n+1} - (4n+2) a_n = -4 B(2n, n). Again set a_n = b_n B(2n, n). Then (4n+2)(b_{n+1} - b_n) = -4 or b_{n+1} - b_n = -1/(2n+1). We then get b_n = -2(1 + 1/3 + 1/5 + ... + 1/(2n-1)) = -(2 H_{2n} - H_n) and so a_n = -(2H_{2n} - H_n) B(2n, n). Combining these we get [2 log((1 + sqrt(1-4y))/2) - log(1-4y)]/sqrt(1-4y) = sum_{n=1}^infinity H_n B(2n,n) y^n and also [log((1 + sqrt(1-4y))/2) - log(1-4y)]/sqrt(1-4y) = sum_{n=1}^infinity H_{2n} B(2n,n) y^n. Robin Chapman + "They did not have proper School of Mathematical Sciences - palms at home in Exeter." University of Exeter, EX4 4QE, UK + rjc@maths.exeter.ac.uk - Peter Carey, http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda, chapter 20 ----- Posted via Deja.com, The People-Powered Information Exchange ----- ------ http://www.deja.com/ Discussions * Ratings * Communities ------