Re: A curious result?

*To*: mathgroup at smc.vnet.net*Subject*: [mg91511] Re: A curious result?*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Sun, 24 Aug 2008 07:08:02 -0400 (EDT)*References*: <g8o89u$l8q$1@smc.vnet.net>

amzoti <amzoti at gmail.com> wrote: > a. Sum[1/n^1.001, {n, 1 , Infinity}] = 1000.58 > > b. Integrate[1/n^(1.001), {n, 1, Infinity}] = 1000 I don't see anything curious about either result. Copied below is my response to what you had posted earlier in sci.math. David ------------------------------------------------- David W. Cantrell <DWCantrell at sigmaxi.net> wrote: > amzoti <amzoti at gmail.com> wrote: > > Hi All, > > > > I just finished reading this nice write-up on precision that be > > required reading for all students. > > > > www.stewartcalculus.com/data/default/upfiles/LiesCalcAndCompTold.pdf > > > > Which one of these is correct? > > Both. > > > a. Sum[1/n^1.001, {n, 1, Infinity}] = 1000.58 > > You appear to be using Mathematica. Note that it is not a good idea, > typically, to combine an "approximate" number (in this case, 1.001) with > a symbolic operation (in this case, Sum). Much better would be to use, > say, > > Sum[1/n^(1001/1000), {n, 1, Infinity}] > > which returns the precise result > > Zeta[1001/1000] > > and that can then be approximated, if desired. Rounded to two decimal > places, 1000.58 is correct. > > > b. Integrate[1/n^(1.001), {n, 1, Infinity}] = 1000 > > Well, actually that's not what Mathematica gives for the result. Since > the input involved an "approximate" number, the result is 1000. instead. > > But, as for part a., it would have been much better to have asked for > something which is exact: > > Integrate[1/n^(1001/1000), {n, 1, Infinity}] > > yields 1000 precisely. > > David