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Topic: Can two series, both diverges, multiplied give a series that converges?
Replies: 22   Last Post: Oct 7, 2017 12:52 AM

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 Karl-Olav Nyberg Posts: 1,575 Registered: 12/6/04
Re: Can two series, both diverges, multiplied give a series that converges?
Posted: Oct 6, 2017 1:00 PM

fredag 6. oktober 2017 18.05.49 UTC+2 skrev Mike Terry følgende:
> On 06/10/2017 16:21, konyberg wrote:
> > fredag 6. oktober 2017 17.13.31 UTC+2 skrev Peter Percival følgende:
> >> 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
> >>>

> > It is the multiplication of the two series.
>
> That doesn't answer Peter's question. Each series has infinitely many
> terms, and you need to say what you mean the product to be calculated
> from those terms.
>
> If you thought this through carefully, you'd realise straight away the
>
> To get you started in the right direction, suppose the first series is:
>
> Sum [n=1 to oo] (a_n)
>
> and the second is:
>
> Sum [n=1 to oo] (b_n)
>
> Now, what do you mean by the "product" of these series?
>
> If you feel tempted to reply "just multiply them together", then ask
> yourself "multiply WHAT together exactly?" (Remember, multiplication is
> an operation that takes TWO numbers, and gives a single number as the
> answer. In the two series, you have INFINITELY many numbers...)
>
> Or perhaps your answer will be that the product of the two series is
> some new third series? (If so, then say what is the n'th term of this
> new series?)
>
>
> Regards,
> Mike.

There are two ways to give the product:
1: sum(a) * sum (b) = sum (c) or
2: every of sum a multiplied with the ones of sum of b.
If the sum of any of them is not (in 2) as we like, then none of them are!
KON