```Date: Oct 6, 2017 11:53 AM
Author: Karl-Olav Nyberg
Subject: Re: Can two series, both diverges, multiplied give a series that converges?

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 FanI 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)?KONKON
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