The Math Forum



Search All of the Math Forum:

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


Math Forum » Discussions » sci.math.* » sci.math

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Mike Terry

Posts: 763
Registered: 12/6/04
Re: Can two series, both diverges, multiplied give a series that
converges?

Posted: Oct 6, 2017 4:30 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 06/10/2017 17:24, konyberg wrote:
> fredag 6. oktober 2017 18.17.41 UTC+2 skrev konyberg følgende:
>> 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
>>> answer to your original question, I think!
>>>
>>> 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.

>>
>> Or consider my first is sum(a) is like sum(b), where both sum(a) and sum(b) goes to inf. What is then the product of them?
>> KON

>
> Or on the other side:
> One a goes to inf, the other goes to 0. What is now the product?
> You can not tell if you do not know the functions defining the entities.
> KON
>


My best advice to you, is that if questions like these interest you, you
need to find an introductory book on "Real Analysis". Such a book would
rigorously define what is meant by "sequence", "series", "convergence",
"absolute convergence" etc., which are needed to understand questions
like these. (It is too much to be taught all of this through a forum
like sci.math!)

Still, I'll have a go at sketching out an answer to what I think you're
really asking...

[Blimey! Having finished finally I see I've ended up writing a "War and
Peace" novel. Or we could say I've "done a Jim Burns!!" which I mean in
a positive way, in case he reads this :)]


First, bear in mind that a series is written to look like an infinite
sum, but in fact we cannot actually perform an infinite sum, only finite
sums. So if we have an infinite SERIES like this:

a_1 + a_2 + a_3 + a_4 ...

we first translate it to a corresponding SEQUENCE, consisting of the
partial sums, i.e. the sequence (A_n) where

A_1 = a_1
A_2 = a_1 + a_2
A_3 = a_1 + a_2 + a_3
A_4 = a_1 + a_2 + a_3 + a_4
...
then we ask whether the sequence (A_1, A_2, ...) converges. If it
converges to a real number A, then we say the series Sum(a_n) converges
to A. (If (A_n) doesn't converge we say the series diverges.)

So suppose we have a second series Sum(b_n), and corresponding
partial-sum sequence B_n.

You want to multiply them together, but you're not sure what this means.

Let's start simply by assuming all the a_n and b_n terms are
non-negative real numbers, and both the series converge.

a_1 + a_2 + a_3 + a_4 ... = lim[n->00] A_n = A

b_1 + b_2 + b_3 + b_4 ... = lim[n->00] B_n = B

whatever you mean by "multiplying the series", I suppose you will want
the final result to correspond to the product of the limits A and B??

This seems reasonable for me to assume, but not a given, because in the
end a series is a DEFINED mathematical concept, and there is no built in
meaning to the "product" or "sum" etc. of two such series until we
define what we mean by those terms. Of course, we would want to make
the definition intuitive, and IF POSSIBLE ensure that at least with the
above example that the "product" of the series evaluates to A*B.

So... what about just multiplying A and B, and saying that's what we
mean by the product of the two series? We can do that, and it sort of
ties up with your "just multiply them together" phrase, so you could be
thinking of this. Personally, I would say that this is more a case of
multiplying two numbers [the two SUMS of the series] rather than
multiplying the series themselves - at least, there's no new idea
involved, doing it like this.

OK then, given the two series, to get their product we first sum each
series, then multiply those numbers, no problem. :) If we stsick with
this, your original question is easy!

Sum(1/(n)) diverges

Sum(1/(1+n)) diverges

so we don't have an A and B to multiply, so we can't multiply the two
series. No problem! You said in your original post "we can argue that
every product of the new series is smaller or equal to 1/n^2" but that
makes no sense with this approach - there is no "every product" to
consider in the first place!

So... you are hoping for some other definition for the product of two
series :)

Sorry for meandering a bit, but I think it would be instructive here to
quickly look at the corresponding question of ADDING two series rather
than MULTIPLYING them, because it is much simpler, and shows some of the
same issues...

As with above we start with non-negative, convergent series:

a_1 + a_2 + a_3 + a_4 ... = lim[n->00] A_n = A

b_1 + b_2 + b_3 + b_4 ... = lim[n->00] B_n = B

and we wonder how we can define the SUM of these two series.

Following the simple approach above of just saying the SUM is A+B is
fair enough, but you can probably see there's a totally different
approach we might consider: what happens if we form a THIRD series
Sum(c_n) each of whose terms is just the sum of the corresponding terms
from the original series?

c_1 = a_1 + b_1
c_2 = a_2 + b_2
...

Then we get the following new series:

c_1 + c_2 + c_3 + c_4 ... = lim[n->00] C_n = ????

and we could say that by the "SUM" of the two original series Sum(a_n),
Sum(b_n), we mean BY DEFINITION the new series Sum(c_n).

The main point is that this is COMPLETELY different from just saying
that by the "SUM" we mean by definition A+B. Hopefully this is obvious,
but it's why Peter and I wanted you to say what you intended by the
phrase "multiply the two series" - there are multiple possible answers
and so we must choose/define what such a phrase means!

When people talk about adding two infinite series, they generally mean
the approach above of forming the new series Sum(c_n) = Sum(a_n + b_n),
not just adding A and B.

But now we have a new worry here - can we say that if Sum(a_n) and
Sum(b_n) both converge, say to A and B respectively, that the "sum of
these series" as we've DEFINED it, i.e. Sum(c_n) also converges, say to
C, and that we will always have C = A+B ?

Don't make the mistake of thinking this is "automatically true by
definion"! We have DEFINED instead we would need to prove this as a
theorem. The good news is that it is quite easy to prove this theorem,
so everything works out nicely!

So now let's return to the PRODUCT of two infinite series - can we do a
similar approach here, and somehow define the product of two series to
be a third (new) series whose terms are somehow related back to the
terms in the original series? [I'm pretty sure this is basically what
you're looking for here!]

The answer is "yes, and no - things aren't so simple for products, as
compared to the case with sums!". :(

So for example, copying exactly the idea c_n = a_n + b_n and just
changing the + to a * doesn't work! There's no problem with defining a
new series where c_n = a_n * b_n, but the limit of the resulting series
won't in general be the product of the limits of the two original
series. We can see even with finite sums this approach is hopelessly
wrong! E.g. with simple 3-term series

(2 + 1 + 3) (1 + 5 + 7) = 6*13 = 78, whereas

(2*1 + 1*5 + 3*7) = 2 + 5 + 21 = 28 (not even close!)

So no chance of this working for infinite sums :)

Looking at WHY the multiplying of (2+1+3)(1+5+7) = 78 failed when we
just multiplied "corresponding" terms and added, you'll hopefully
realise the reason is we missed out all the "cross terms" - i.e. as well
as (2*1 + 1*5 + 3*7) there need to be all the cross-terms such as 2*5
2*7, 1*1 and so on. (That's basic algebra which I expect you know.)

Can we do something similar with multiplying two series, so that we
include all the cross-terms? This is where the answer is a guarded
"yes", but things get much more complicated than when we were just
looking at sums :( To make this work, when trying to multiply two
series Sum(a_n) and Sum(b_n), we need to consider every single
cross-term product of the form (a_n * b_m) where here n and m are
independent indexes that together range over all the combinations of
ordered pairs of natural numbers.

We can show all the cross-products in an infinite table like this:
[use fixed width font to keep things lined up!]

--+-------------------------------------------------- ...
| a_1*b_1, a_1*b_2 a_1*b_3 a_1*b_4, ....
|
| a_2*b_1, a_2*b_2 a_2*b_3 a_2*b_4, ....
|
| a_3*b_1, a_3*b_2 a_3*b_3 a_3*b_4, ....
|
| ...

[the table runs on forever heading towards the right, and heading
downwards, but has a definite top line and a left-hand column]


Can we get a new series Sum(c_n) out of this? Yes! The most common way
is to take each "diagonal" (running in direction "upwards+rightwards")
and make a new term c_n by adding the finite number of elements on that
diagonal, and then Sum(c_n) is our new series. In other words:

c_1 = a_1*b_1
c_2 = a_2*b_1 + a_1*b_2
c_3 = a_3*b_1 + a_2*b_2 + a_1*b_3
c_4 = a_4*b_1 + a_3*b_2 + a_2*b_3 + a_1*b_4
... etc.

(Note: each term c_n is the sum of only finitely many terms, but as n
increases we get more and more of these terms...)

This new series is called the "Cauchy product" of the original two series.

So finally I'm thinking this is probably the best answer (from your
perspective) to the question of how we can define the product of two
infinite series, as a new series. But see how much more complicated
this is compared with the situation when we were just ADDING two series!

In the case of adding two series we were able to simply define
c_n = a_n + b_n
then there is a simple theorem to help us, which says basically that if
the two original series converge, then Sum(c_n) also converges, and it
converges to the sum of the two sums of the original series. There are
no additional conditions or requirements that have to be satisfied, so
everything runs smoothly.

For multiplying two series, we COULD define a simple new series
c_n = a_n * b_n
but we've seen that doesn't work out as we want. Or, we can define the
Cauchy product series, but this is a bit trickier even to write down,
and worse is to come - we don't have the nice *SIMPLE* theorem we want
to tell us "this approach works". I.e. basically we want a theorem that
assures us that after doing all this, we will have

Sum(c_n) = Sum(a_n) * Sum(b_n)

[Remember that in the above expression, each of the Sum() terms is a
separate limit in its own right. So the equation is far from obviously
true!]

None the less, there are conditions under which the required theorem
holds. Specifically, in the case that Sum(a_n) and Sum(b_n) both
converge, and ONE OF THEM CONVERGES ABSOLUTELY. [look it up!]

With your example a_n = 1/n, b_n = 1/(n+1), all the terms are positive,
which is helpful, however, both Sum(a_n) and Sum(b_n) diverge... Hmmm,
now I need to think!

OK, it can be proven that if (as in your example) all the terms are
positive, and both series diverge, then the Cauchy product series c_n
will also diverge. (That's fairly easy to prove.)

So what about your original sentence "we can argue that every product of
the new series is smaller or equal to 1/n^2" ?

Actually this is a reasonable question! :) I think the main problem
with this is that there are a lot more terms to be added in the product
than you were thinking! Let's look at the terms in a particular
diagonal that make up one of the c_n. E.g. c_5 for your example:

c_5 = a_5*b_1 + a_4*b_2 + a_3*b_3 + a_2*b_4 + a_1*b5

= (1/5)(1/2) + (1/4)(1/3) + (1/3)(1/4) +(1/2)(1/5) +(1/1)(1/6) +

It's clear that middle term of this sum is (1/3)(1/4), which is less
than 1/3^2, so we can say as n increases it is "of the order of" 1/n^2.
(And yes, Sum(1/n^2) converges...)

The thing being overlooked is this is just one term making up c_n. Even
if the other terms were all like 1/n^2 as well, the number of terms
would be growning as n, so the effect that would make c_n grow like 1/n.
(However, clearly each c_n contains one term (the rightmost term)
which is exactly 1/(n+1), and already that 1 term is enough to make
Sum(c_n) diverge!

HTH,
Mike.




Date Subject Author
10/6/17
Read Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Markus Klyver
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that
converges?
Peter Percival
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that
converges?
Mike Terry
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
bursejan@gmail.com
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
bursejan@gmail.com
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that
converges?
Mike Terry
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/7/17
Read Re: Can two series, both diverges, multiplied give a series that
converges?
William Elliot
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
bursejan@gmail.com
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
bursejan@gmail.com
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
bursejan@gmail.com
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Jan Bielawski
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that converges?
Karl-Olav Nyberg
10/7/17
Read Re: Can two series, both diverges, multiplied give a series that
converges?
William Elliot
10/6/17
Read Re: Can two series, both diverges, multiplied give a series that
converges?
William Elliot

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2017. All Rights Reserved.