# Thread: Infinite product of a sequence of real numbers

1. ## Infinite product of a sequence of real numbers

I'm not sure what this is defined as, so I'm putting this here for lack of better knowledge.

I'm working on problems involving the infinite product of a sequence of real numbers, as well as a couple of proofs involving the infinite product of a sequence of real numbers. However, my current problem deals with solving the infinite product part itself.

I understand that you're taking the limit of the infinite product to get the result, but I don't understand how to GET the limit of an infinite product. Can anyone point me in the right direction(s)?

Real Analysis is getting to me.

2. Originally Posted by Jedi Arashi
I'm not sure what this is defined as, so I'm putting this here for lack of better knowledge.

I'm working on problems involving the infinite product of a sequence of real numbers, as well as a couple of proofs involving the infinite product of a sequence of real numbers. However, my current problem deals with solving the infinite product part itself.

I understand that you're taking the limit of the infinite product to get the result, but I don't understand how to GET the limit of an infinite product. Can anyone point me in the right direction(s)?

Real Analysis is getting to me.
Consult The Force, young Jedi

if the force does not help, a Jedi Master here will

3. That's what I get for being a sci-fi dork.

I tried Wiki, but I'm still not quite getting it (guess I'm kind of an idiot). Does it have to do with e, which is what I saw one source say?

4. Originally Posted by Jedi Arashi
That's what I get for being a sci-fi dork.

I tried Wiki, but I'm still not quite getting it (guess I'm kind of an idiot). Does it have to do with e, which is what I saw one source say?
as you can see from The Force, it has to do with logs and e ("exp" means e)

perhaps it would be better if you posted a particular problem. i do not know enough about this topic to explain it in general. if that does not feed your desire to excel, you can wait for one of the other Jedi Masters to assist you, young Padawan. (i hope i spelled that right)

5. I commend you, as you did spell it correctly. Congrats--I know fans of the series who can't even spell that.

I could post one of the problems, but I'm not the greatest with LaTeX, so it'll have to be plain text...?

The infinite product from k = 1 to infinity of (1 - (1/k) ) = ? It seems to me that it would be zero, from doing about one hundred iterations on my calculator (hush!), but I need to show the "work". The other problem that wasn't a proof was the infinite product from k = 1 to infinity of (1 - (1/(k^2))), which also seems to be zero. Am I anywhere near on track?

6. Originally Posted by Jedi Arashi
I commend you, as you did spell it correctly. Congrats--I know fans of the series who can't even spell that.

I could post one of the problems, but I'm not the greatest with LaTeX, so it'll have to be plain text...?

The infinite product from k = 1 to infinity of (1 - (1/k) ) = ? It seems to me that it would be zero
i fear you are correct. but don't take my word for it. i have little experience with infinite products...ok, no experience. but here's what i figured from The Force, maybe it can point you in the right direction since you are actually in a class where you're doing this stuff and would know more about the topic than i would:

we wish to find $\displaystyle \prod_{k = 1}^{\infty} \left( 1 - \frac 1k \right)$

Now, $\displaystyle \ln \prod_{k = 1}^{\infty} \left( 1 - \frac 1k \right) = \sum_{k = 1}^{\infty} \ln \left( 1 - \frac 1k \right)$ since the log of products is the sum of the log of each term (recall $\displaystyle \log ab = \log a + \log b$)

but $\displaystyle \sum_{k = 1}^{\infty} \ln \left( 1 - \frac 1k \right)$ sums infinitely many negative terms, the first of which is pretty much negative infinity, thus:

$\displaystyle \ln \prod_{k = 1}^{\infty} \left( 1 - \frac 1k \right) = \sum_{k = 1}^{\infty} \ln \left( 1 - \frac 1k \right) \approx - \infty$

since $\displaystyle \ln \prod_{k = 1}^{\infty} \left( 1 - \frac 1k \right) = - \infty$

$\displaystyle \Rightarrow e^{- \infty} = \prod_{k = 1}^{\infty} \left( 1 - \frac 1k \right) = 0$

of course, you would use limits to write this more elegantly/rigorously, since actually plugging in $\displaystyle \infty$ into a formula is a no-no. but i think you get the idea

hope it helps, young Jedi

it makes sense though, since the first term in the product is actually zero! so i guess you could just say that. the largest any term can be is 1 (and it won't actually get to 1, 1 is the limit of the terms) and the first term is zero, so obviously the products would be zero. it may be a little more interesting if we started at k = 2 or something

7. Crud, it was from k = 2 to infinity for both problems. Oops!

But yeah, it's something like that, I think...I took the product of the limits, and some terms canceled, leaving you with the limit of either 1/n or 1/(n^2), which are obviously both zero, so...I think I have that much.

Now it's the proofs, which are...well...hard. I'd post them here, but the problem remains that the assignment is due tomorrow, and I couldn't go in for help from the teacher earlier due to health problems. I seriously hope I pass this class.

Thanks for the help!

8. Originally Posted by Jedi Arashi
Crud, it was from k = 2 to infinity for both problems. Oops!

But yeah, it's something like that, I think...I took the product of the limits, and some terms canceled, leaving you with the limit of either 1/n or 1/(n^2), which are obviously both zero, so...I think I have that much.
i see what you are saying. i wrote it out, and it sounds good to me. (i turned $\displaystyle 1 - \frac 1k$ into $\displaystyle \frac {k - 1}k$ first, which is what i suppose an honorable Jedi like you did)

Now it's the proofs, which are...well...hard. I'd post them here, but the problem remains that the assignment is due tomorrow, and I couldn't go in for help from the teacher earlier due to health problems. I seriously hope I pass this class.

Thanks for the help!
proofs? you have to write proofs for these? they seem like really computational problems, why are proofs needed?

9. The proofs aren't on those computational ones. The proofs involve...hang on, let me grab my book (trying to do math with a half-sleeping puppy on one's lap can be difficult).

The first proof is, "If the infinite sum of a sub k from k = 1 to infinity equals p, with p not zero, prove that the limit of a sub k as k goes to infinity is 1".

I'm so lost. I get that, apparently, if it's not zero, and it's convergent, then generally the product will equal one, but I'm lost as to how to PROVE it.

We never even went over these in class, per se, they were just tacked on to the end of a chapter on limits and sequences of real numbers and the like.

10. Originally Posted by Jedi Arashi
The proofs aren't on those computational ones. The proofs involve...hang on, let me grab my book (trying to do math with a half-sleeping puppy on one's lap can be difficult).

The first proof is, "If the infinite sum of a sub k from k = 1 to infinity equals p, with p not zero, prove that the limit of a sub k as k goes to infinity is 1".

I'm so lost. I get that, apparently, if it's not zero, and it's convergent, then generally the product will equal one, but I'm lost as to how to PROVE it.

We never even went over these in class, per se, they were just tacked on to the end of a chapter on limits and sequences of real numbers and the like.
I suppose you meant to say $\displaystyle \prod_{k = 1}^{\infty}a_k = p \ne 0$

Now, we know that $\displaystyle \ln \prod_{k = 1}^{\infty}a_k = \sum_{k = 1}^{\infty}\ln a_k = p$

now, if the series converges, we must have that $\displaystyle \lim_{k \to \infty} \ln a_k = 0$

So, $\displaystyle \lim_{k \to \infty} \ln a_k = \ln \left( \lim_{k \to \infty} a_k \right) = 0$

but this happens if and only if $\displaystyle \lim_{k \to \infty}a_k = 1$, since $\displaystyle \ln 1 = 0$

QED

11. Yes, forgive my typo.

Your proof makes sense, really, and this always happens when someone goes through a proof with me. I can never get how to start one, but usually if someone gives me a push in the right direction, it all sort of falls into place all of a sudden.

I'm not sure why I'm having trouble understanding the bit with the ln. They never mentioned the stupid thing in the book, but it makes sense when you put it like that.

There's another proof, which I can post tonight, but I don't have my book with me now.

Thanks again.

12. Originally Posted by Jedi Arashi
Yes, forgive my typo.

Your proof makes sense, really, and this always happens when someone goes through a proof with me. I can never get how to start one, but usually if someone gives me a push in the right direction, it all sort of falls into place all of a sudden.

I'm not sure why I'm having trouble understanding the bit with the ln. They never mentioned the stupid thing in the book, but it makes sense when you put it like that.
yeah, i'm like that with a lot of things. the ln thing is just a trick though. since we know "a lot" of stuff about series, it is usually easier working with an infinite series than an infinite product. since logarithms turn products into sums, we can apply the logarithms to an infinite product to get infinite sums (and you can use induction to show that this is valid).

There's another proof, which I can post tonight, but I don't have my book with me now.

Thanks again.
i'd like to see it when you get the time--and the book

13. All right, I've got my book. I had a bit of a time finding this thread once they moved it (how did I not see the "real analysis" bit on this forum?). Here's the other proof in plain text:

If a sub n is greater than or equal to zero for all n that exists in the Natural Number set, prove that the infinite product from k = 1 to infinity of (1 + a sub k) converges if and only if the infinite sum from k = 1 to infinity of a sub k converges. To prove the result, establish the following inequality: a sub 1 + ... + a sub n <= (1 + a sub 1) * .... * (1 + a sub n) <= e^(a sub 1 + .... + a sub n).

Does that make any sense? I get that it's true and why it's true in a sort of "plain English" way, but proving it (and it being an if and only if statement at that) is giving me issues.