Re: Proving inequalities with Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg52499] Re: [mg52491] Proving inequalities with Mathematica*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Tue, 30 Nov 2004 05:24:01 -0500 (EST)*References*: <200411290622.BAA27997@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 29 Nov 2004, at 15:22, Toshiyuki (Toshi) Meshii wrote: > Hi, > > I was wondering whether Mathematica is useful for proving a problem of > inequality. > My problem is as follows: > > Let An and Bn (n=1,2,3...) be real sequences. > Some characteristics of these sequences are known. > > i) Abs[An+1/An] < 1 > ii) Abs[Bn+1/Bn] < 1 > iii) Sum[An, {1,Infinity}] = 0 > iv) Sum[An*Bn, {1, Infinity}] = alpha (note: a real number) > > Then I want to prove with Mathematica that > 0 < Abs[alpha] < Abs[A1*B1] > > Does anyone have an idea? > > -Toshi > First of all, you have written your condition as > Abs[An+1/An] < 1 Presumably you meant n+1 to be a subscript (if not the condition is always false). Assuming that, Mathematica can certainly be helpful in proving your claim to be false. Here is an example. We define the sequence A[n] as follows: A[1] = -1; A[2] = Log[2]; A[n_] = (-1)^(n - 1)/(n - 1); You can check that the condition Abs[A[n+1]/A[n]]<1 is always satisfied. We also define the sequence B[n] by B[i_] := A[i] so that the condition Abs[B[n+1]/B[n]]<1 is satisifed. We check that (iii) is satisfied: A[1]+A[2]+Sum[A[n],{n,3,Infinity}]//Simplify 0 Let's now compute alpha: alpha=A[1]*B[1]+A[2]*B[2]+Sum[A[i]*B[i],{i,3,Infinity}]//N 2.12539 Let's see if your inequality holds: Abs[alpha]<Abs[A[1]*B[1]] False Oops ... Andrzej Kozlowski Chiba, Japan http://www.akikoz.net/~andrzej/ http://www.mimuw.edu.pl/~akoz/

**References**:**Proving inequalities with Mathematica***From:*"Toshiyuki \(Toshi\) Meshii" <meshii@mech.fukui-u.ac.jp>