Re: I broke the sum into pieces

*To*: mathgroup at smc.vnet.net*Subject*: [mg105089] Re: I broke the sum into pieces*From*: Roger Bagula <rlbagula at sbcglobal.net>*Date*: Fri, 20 Nov 2009 06:40:50 -0500 (EST)*References*: <4B0551F8.1050308@sbcglobal.net> <144987c90911190624s3f42b7e7g6545e37062631d1f@mail.gmail.com>*Reply-to*: rlbagulatftn at yahoo.com

Alexander Povolotsky wrote: > Did you verify that with your friend - that guy who runs Mathematica group ? > On 11/19/09, Roger Bagula <rlbagula at sbcglobal.net> wrote: > >> Alexander Povolotsky >> >> This is the part Mathematica fails on: >> Sum[(1/(64^n))*(1/(32*n + 333333)), {n, 0, Infinity}] >> for some reason. >> In[36]:= >> N[333333/64,100] >> >> Out[36]= >> 5208.3281250000000000000000000000000000000000000000000000000000000000000000000\ >> 00000000000000000000000 >> >> In[39]:= >> b[5]=Sum[(64^(-n-1))*(1/(n/2+5208.328125)),{n,0,Infinity}] >> >> Out[39]= >> \!\(\(-2.05555574498401916904809250313`15.954589770191003*^18796\)\) >> It just gives the wrong answer for that part. >> Respectfully, Roger L. Bagula >> 11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : >> http://www.google.com/profiles/Roger.Bagula >> alternative email: roger.bagula at gmail.com >> >> >> >> > > Alexander Povolotsky, No I haven't reported this. Since the Mathematica guys ( Trott and Wolfram himself) put up my Beta cube fractal without giving me any credit for it, they are on my list, ha, ha... In trying to get a 5th term I tried: $MaxExtraPrecision = 200 b[5] = Sum[(1/(64^n))*(1/(32*n + 3*1097)), {n, 0, Infinity}]; N[%, 100] 0.0003086349990279674310845177265991859681618864099951680507521205061977799618\ 089309382536127817971331870245951164951`99.99999999999999 I'm pretty sure that is wrong too. I'll cc this to him. I can't use this approach to get closer to Pi at this rate. Clear[b] b[1]=Sum[(1/(64^n))*22/(32*n+7),{n,0,Infinity}]; b[2]=Sum[(1/(64^n))*(-1/(32*n+114)),{n,0,Infinity}]; b[3]=Sum[(1/(64^n))*(-1/(32*n+394)),{n,0,Infinity}]; b[4]=Sum[(1/(64^n))*(1/(32*n+781)),{n,0,Infinity}]; Sum[b[m],{m,1,4}]; N[%,100] 3.1415920104768860463819395769548831674912014189966869630288247319187743718795\ 561171365118029140148452540476917807442`99.99999999999999 %-Pi -6.431129071920807038063246197167059679803784188579461198603890420344066528814\ 91523022428102222728100394732538`93.31113744211372*^-7 Respectfully, Roger L. Bagula 11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : http://www.google.com/profiles/Roger.Bagula alternative email: roger.bagula at gmail.com

**Follow-Ups**:**Re: Re: I broke the sum into pieces***From:*Daniel Lichtblau <danl@wolfram.com>