Re: How to show 1+2+3+ ... = -1/12 using

*To*: mathgroup at smc.vnet.net*Subject*: [mg132241] Re: How to show 1+2+3+ ... = -1/12 using*From*: Itai Seggev <itais at wolfram.com>*Date*: Tue, 21 Jan 2014 03:49:15 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <lbg084$am4$1@smc.vnet.net>

On Mon, Jan 20, 2014 at 04:00:45AM -0500, Richard Fateman wrote: > On 1/18/2014 11:52 PM, Matthias Bode wrote: > > Hola, > > > > I came across this video (supported by the Mathematical Sciences Research Institute* in Berkeley, California): > > > > http://www.numberphile.com/videos/analytical_continuation1.html > > > > Could the method shown in this video be replicated using Mathematica symbols such as Sum[] &c.? > > > > Best regards, > > > > MATTHIAS BODES 17.36398=B0, W 66.21816=B0,2'590 m. AMSL. > > > > *) http://www.msri.org/web/msri > > > > Sure. Piece of cake. > Sum[a^n,{n,0,Infinity}] results in 1/(1-a). > > %/. a->-1 tells you that this Sum is 1/2 > > Starting from this lie (the sum is actually divergent), you should > be able to prove lots and lots of things. Or more usefully: In[43]:= Sum[1/n^s, {n, 1, \[Infinity]}] /. s -> -1 Out[43]= -(1/12) Which is perefectly valid if you understand what analytic continuation is. -- Itai Seggev Mathematica Algorithms R&D 217-398-0700