```Date: Oct 6, 2017 12:41 PM
Author: bursejan@gmail.com
Subject: Re: Can two series, both diverges, multiplied give a series that converges?

You find this product of two infinite seriesin these notes here:    MATH 304: CONSTRUCTING THE REAL NUMBERS,   Peter Kahn Spring 2007    http://www.math.cornell.edu/~kahn/reals07.pdfSee page 12, 4.2 Algebraic operations on sequencesthere it is defined the operation (*) between twosequences (of partial sums) as follows:     c. {an} * {bn} = {an * bn}But if your partial sums are really 1/n and 1/(n+1)then the result is different. You wrote 1/n and 1/(n+1)that these are the summands?In the case the series had as partial sums 1/n,the summands would have the negative value:     1/(k+1)-1/kThis will give as partial sums:    1    1/2  = 1 + (1/2-1)    1/3  = 1 + (1/2-1) + (1/3-1/2)    1/4  = 1 + (1/2-1) + (1/3-1/2) + (1/4-1/3)    Etc..Or differently expressed:    1    1/2  = 1 - 1/2    1/3  = 1 - 1/2 - 1/6    1/4  = 1 - 1/2 - 1/6 - 1/12    Etc..Could you be more specific what your series are?Ok you are specific in your original post, thereyou write s = lim (n=1 to inf) Sum(1/n) and t = lim (n=1 to inf) Sum(1/(1+n)). So the product diverges and doesn't give pi^2/6 or some such.Am Freitag, 6. Oktober 2017 18:32:30 UTC+2 schrieb burs...@gmail.com:> One approach is to use {sn*tn} for the product of {sn}> and {tn}. I dont know what the closed form is for the> case when sn and tn are as follows:> >      sn = sum_i=1^n 1/i>   >      tn = sum_i=1^n 1/(i+1)> > Since I am lazy I asked Wolfram Alpha for help. > The new series will be {sn*tn} or lets call this> seies {rn}, we will have for the partial sums simply:>  >      rn = sn*tn> >         = (sum_i=1^n 1/i)*(sum_i=1^n 1/(i+1))> > Note the above definition of rn is used in the> construction of the reals from Cauchy series. Anyway> the resulting series in closd form is> >     rn = H_n (H_(n + 1) - 1)> > And the limit is:> >    lim n->oo rn = oo> > Am Freitag, 6. Oktober 2017 17:53:58 UTC+2 schrieb konyberg:> > fredag 6. oktober 2017 17.32.52 UTC+2 skrev burs...@gmail.com fĂ¸lgende:> > > For two series {sn} and {tn} just define> > > the product series as {sn*tn}, note this is NOT:> > > > > >      sum_i^oo ai*bi      /* NOT the product */> > > > > > where sn=sum_i^n ai and tn=sum_i^n bi. The> > > above is also not the Cauchy product.> > > > > > Here is the product of two harmonic series:> > > > > > sn      sn^2> > > 1	1> > > 1.5	2.25> > > 1.833333333	3.361111111> > > 2.083333333	4.340277778> > > 2.283333333	5.213611111> > > 2.45	6.0025> > > 2.592857143	6.722908163> > > etc...> > > > > > It pretty much diverges, since sn < sn^2,> > > and sn diverges. And its not sum 1/n^2.> > > > > > Am Freitag, 6. Oktober 2017 17:13:31 UTC+2 schrieb Peter Percival:> > > > konyberg wrote:> > > > > Consider these two series. s = lim (n=1 to inf) Sum(1/n) and t = lim> > > > > (n=1 to inf) Sum(1/(1+n)). Both series diverges, going to infinity.> > > > > Now if we multiply these,> > > > > > > > What is the definition of the product of two infinite series?> > > > > > > > > > > > > we can argue that every product of the new> > > > > series is smaller or equal to 1/n^2. So it should converge. Or can> > > > > we? Let us write the first as a series without the sigma and the> > > > > other with sigma. s*t = (1+1/2+1/3+ ...) * t. And since the first> > > > > from s (1 * t) diverges, how can s*t converge?> > > > >> > > > > KON> > > > >> > > > > > > > > > > > -- > > > > Do, as a concession to my poor wits, Lord Darlington, just explain> > > > to me what you really mean.> > > > I think I had better not, Duchess.  Nowadays to be intelligible is> > > > to be found out. -- Oscar Wilde, Lady Windermere's Fan> > > > I know that using the definitions of these two series, and operating on them is not going to work. However. Can you give what the two series, when multiplied gives? The series have the sequence 1/n and 1/(n+1), what is these two multiplied? I do know how to give the two different series (and also the product of these). But how do you multiply two infinite series (by hand)?> > KON> > KON
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