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Topic: Can two series, both diverges, multiplied give a series that converges?
Replies: 3   Last Post: Oct 6, 2017 2:16 PM

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Karl-Olav Nyberg

Posts: 1,517
Registered: 12/6/04
Re: Can two series, both diverges, multiplied give a series that converges?
Posted: Oct 6, 2017 1:48 PM
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fredag 6. oktober 2017 19.36.54 UTC+2 skrev John Gabriel følgende:
> On Friday, 6 October 2017 13:31:46 UTC-4, konyberg wrote:
> > fredag 6. oktober 2017 19.24.37 UTC+2 skrev John Gabriel følgende:
> > > On Friday, 6 October 2017 13:20:01 UTC-4, John Gabriel wrote:
> > > > On Friday, 6 October 2017 11:53:58 UTC-4, konyberg wrote:
> > > > > fredag 6. oktober 2017 17.32.52 UTC+2 skrev burs...@gmail.com følgende:
> > > > > > 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?

> > > >
> > > > Very stupid question. The product is given by the product of the general terms, that is, 1/(n(n+1)). This does not converge to the same limit and in two cases, it diverges.
> > > >
> > > > Typically, we talk about convergence as the independent variable increases or decreases without bound and it happens on the entire domain, not just parts of it.
> > > >

> > > > > 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

> > >
> > > But in the case of the product between the two series to infinity, it means multiplying all the terms of one series with the other.

> >
> > Currect,'And how do you do, <nd get the different sums correct and add them?

>
> You CANNOT do unless both of them converge in which case you can conclude that the product will also converge. This is a theorem. And you call yourself a mathematician? Chuckle.


Just to be sure what you mean. A construction like (1 + sqrt(2))m is impossible?
KON



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