# Thread: is this series of products zero?

1. ## is this series of products zero?

Hi,

Given the following two certainties:

a and b are real numbers and may be positive or negative.
Would it be possible to prove the following:

I tried product rule for limits and several other tricks, but I can't do it.
Nonetheless, intuitively this limit seems to be zero. For every a or b in
an infinite set there will be one with opposite sign (otherwise the sums
of a and b are not zero, but they always are), in fact each of the a or b
itself already occurs an infinite number of times. This means that also
every product ab occurs an infinite number of times, with both signs,
or the set would not be infinite. Am I right? Infinity is tricky business...
Is there a basic proof?

Thank you for any help...

2. Welcome to MHF!

1) For what $\displaystyle N$ is the equality true (in the attached file)? Any such N, a big enough one, or a specific one?

2) Have you seen the comparison test for infinite series?

3. Originally Posted by joland
Hi,

Given the following two certainties:

a and b are real numbers and may be positive or negative.
Would it be possible to prove the following:

I tried product rule for limits and several other tricks, but I can't do it.
Nonetheless, intuitively this limit seems to be zero. For every a or b in
an infinite set there will be one with opposite sign (otherwise the sums
of a and b are not zero, but they always are), in fact each of the a or b
itself already occurs an infinite number of times. This means that also
every product ab occurs an infinite number of times, with both signs,
or the set would not be infinite. Am I right? Infinity is tricky business...
Is there a basic proof?

Thank you for any help...

Take $\displaystyle \{a_n\}=\{1,-1,1,-1,\ldots\}\,,\,\,\{b_n\}=\{-1,1,-1,1,\ldots\}$. Then, for some values of $\displaystyle N\in\mathbb{N}$ ,

$\displaystyle \sum\limits^N_{k_1}a_k=\sum\limits^N_{k=1}b_k=0$ , but $\displaystyle a_nb_n=-1\,,\,\forall n\in\mathbb{N}$ ...

Tonio

4. ## thank you

Thank you very much for the response.

My numbers a and b represent deviations with respect to an average value, so the global sum over all "a" (or all "b") is always zero, even for N = 1, N = 2, etc. In your counter example this is not the case, because if the first and last elements of the sets have the same sign, their overall sum is not zero. Only if the signs of the first and final elements are opposite (i.e. there are as many "+1" as "-1" in the sets) the sum of all "a" or "b" is zero. This does not really matter, because the sum of the products aj * bj will never be zero, in fact it will be +-N.

So, your counter example shows that I can not assume that this limit is zero. This is annoying! I was really hoping that this term would cancel out in my (much larger) equation. Back to the drawing board, I guess...

5. Originally Posted by joland
Hi,

Given the following two certainties:

a and b are real numbers and may be positive or negative.
Would it be possible to prove the following:

I tried product rule for limits and several other tricks, but I can't do it.
Nonetheless, intuitively this limit seems to be zero. For every a or b in
an infinite set there will be one with opposite sign (otherwise the sums
of a and b are not zero, but they always are), in fact each of the a or b
itself already occurs an infinite number of times. This means that also
every product ab occurs an infinite number of times, with both signs,
or the set would not be infinite. Am I right? Infinity is tricky business...
Is there a basic proof?

Thank you for any help...
An example: let be...

$\displaystyle a_{n}=\left\{\begin{array}{ll}-1 ,\,\,n=1 \\{}\\\frac{1}{n\ (n-1)} = \frac{1}{n-1} - \frac{1}{n} ,\,\, n>1\end{array}\right.$

$\displaystyle b_{n}=\left\{\begin{array}{ll}1 ,\,\,n=1 \\{}\\\- \frac{1}{n\ (n-1)} = \frac{1}{n} - \frac{1}{n-1} ,\,\, n>1\end{array}\right.$ (1)

It is evident that...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} a_{n} = -1 + 1 -\frac{1}{2} + \frac{1}{2} - \frac{1}{3} +... =0$

$\displaystyle \displaystyle \sum_{n=1}^{\infty} b_{n} = 1 -1 +\frac{1}{2} - \frac{1}{2} + \frac{1}{3} -... =0$ (2)

... but also that...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} a_{n}\ b_{n} = -1 -\frac{1}{4} - \frac{1}{36} - \frac{1}{144} -... \ne 0$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Originally Posted by joland
Thank you very much for the response.

My numbers a and b represent deviations with respect to an average value, so the global sum over all "a" (or all "b") is always zero, even for N = 1, N = 2, etc. In your counter example this is not the case, because if the first and last elements of the sets have the same sign, their overall sum is not zero. Only if the signs of the first and final elements are opposite (i.e. there are as many "+1" as "-1" in the sets) the sum of all "a" or "b" is zero. This does not really matter, because the sum of the products aj * bj will never be zero, in fact it will be +-N.

So, your counter example shows that I can not assume that this limit is zero. This is annoying! I was really hoping that this term would cancel out in my (much larger) equation. Back to the drawing board, I guess...
You will have to give a better definition of the sum, then -- the only way that $\displaystyle \displaystyle \sum_{j=0}^{N} a_j = 0 \ \forall N \in \mathbb{N}$ is if $\displaystyle a_n = 0 \ \forall n \in \mathbb{N}$, but that is obviously not the case..
Also, are the first sums always over a countable set of indices?

7. We have two countable sets of values, call them {x} and {y}.
If X is the mean value of the set of x-values (= sum{x} / N) we can define {a} = {x - X} as set of deviations.
Obviously the sum of all a-values is always zero, because any non-zero mean value (= sum{a} / N) has
gone into the X value. Also, the distribution of a and b is Gaussian, not some highly exceptional set
like {1, -1, 1, -1, ...}
For two such sets of deviations a and b, the set of products aj * bj appears in another expression,
and I'm trying to demonstrate that for large sets of values this term also has zero mean (= zero sum).
Intuitively you would think so, because basically both sets a and b represent white noise, and the
product of two white noise signals (...over a large enough set N) should not all at once contain some
non-zero mean signal. I can't work out the maths though, because it's not really my specialist field.

The more I think of it, the more it looks like a statistical problem that will have a statistical answer,
so I may be in the wrong forum... Perhaps I'll check out some statistics sites, or even electronics.
Those guys should have similar problems all the time.

8. It's really hard (at least for me) to understand what's really going on in there.
I think you'll have a much easier time solving this if you rigorously define the problem.

Good luck