Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

 Messages: [ Previous | Next ]
 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 12:46 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.

I see what you mean.
However. The problem is how to construct the series from multiplying the two series. John Gabriel has given such. His evaluation was correct for his example. Does this mean that John Gabrial's construction is correct always?
KON