# Infite products.

1. Dec 3, 2004

### euclid3.14

Hi, if an = (-1)^n-1 /n what is the infinite product of (1+an)

does working out (1+(a2n-1))(1+a2n) help? :uhh:

2. Dec 3, 2004

### shmoe

Yes it does, what did you get? You should then be able to work out the product up to any finite number of even terms, $\prod_{n=1}^{2k}\left(1+a_n\right)$

Can you do a similar thing (though this will be a bit trickier) to work out the product for a finite number of odd terms, $\prod_{n=1}^{2k+1}\left(1+a_n\right)$?

3. Dec 3, 2004

### euclid3.14

Thanks for the responce!!

I got it to eqaul 1 (think that's right!) I take it that this means the infinite product therefore converges to 1?

Have another querry on infinte products. what does $\prod_{n=1}^{n}\left(1-1/(n+1)^2\right)$? converge to? I've write out the seris to get ((n+1)^2-1(n+2)^-1.....) / ((n+1)^2*(n+2)^2....)

As the sum of 1/(n+1)2 converges I know that the infinite product must converge! :grumpy:

4. Dec 3, 2004

### shmoe

Not so fast, you can only conclude that if it converges then it converges to 1. You can't sum it 2 terms at a time unless you know the thing is absolutely convergent. Consider $\sum(-1)^n$. If you look at the partial sum of an even number of terms, you get 0, but you wouldn't say this sum converges based on that.

You'll want to look at the product of an odd number of terms. What is $\prod_{n=1}^{2k+1}\left(1+a_n\right)$ in terms of k? Use the fact that you know what the product of an even number of terms is...

Can you find a nice expression for $\prod_{n=1}^{k}\left(1-1/(n+1)^2\right)$ for any value of k? Try finding the product for k=1,2,3,4,5 and see if you can guess a nice formula (find more than 5 if you need to). Try to prove your formula is correct and use it to find the limit.

5. Dec 3, 2004

### euclid3.14

Doesn't it tend to one as n tends to infinity as you take the module of the series?

There's a chance that i'm confused!

could simplify it to $\prod_{n=1}^{k}\left((n+1)^2 -1)/(n+1)^2\right$ or $\prod_{n=1}^{k}\left(n^2 +2n /n^2 +2n +1\right)$
Code (Text):
k 1 2 3 4 5
3/4 2/3 5/8 3/5
Seems to be tending to 1/2 but how would could i explain that properly?

Cheers you the responce, it's been very helpful!

Last edited: Dec 3, 2004
6. Dec 3, 2004

### shmoe

I did mean 0. Let me be more precise, let $S_k=\sum_{n=1}^{k}(-1)^{n}$. Then for any k we have $S_{2k}=0$, and $S_{2k+1}=1$, which you can see by pairing up the terms. Thus $\lim_{n\rightarrow \infty}S_{2k}=0$ but $\lim_{n\rightarrow \infty}S_{k}$ does not exist since this sequence of partial sums jumps back and forth between 0 and 1.

The moral is you cannot conclude a sequence converges if all you've shown is the convergence of a subsequence. Hence, I suggest you look at the sequence of partial products containing an odd number of terms to make sure it's behaving.

For your first five partial sums, 3/4, 2/3, 5/8, 3/5, 7/12, do you see any relation between them and the correspodning k values? Let me rewrite these five terms as 3/4, 4/6, 5/8. 6/10, 7/12. Can you express these numerators and denominators as a function of k?

7. Dec 5, 2004

### Tsss

That's right that you can't conclude about the convergence of a sequence only working on his subsequences....in the most cases.
Indeed, if two subsequences has two different limits, then the sequence has no limit (because if a sequence has a limit, then all its subsequences have the same limit).
And, a very useful tip. If $$(U_{2k})$$ and $$(U_{2k+1})$$ have the same limite then $$(U_{k})$$ too.
And that's still right if $$(U_{3k})$$,$$(U_{3k+1})$$,$$(U_{3k+2})$$ converge.

Back to the problem of this thread :).
$$\prod_{k=1}^{n} 1-\frac{1}{(k+1)^{2}}=\prod_{k=1}^{n} \frac{(k+1)^{2}-1}{(k+1)^{2}}=\prod_{k=1}^{n} \frac{k+2}{k+1}\times\frac{k}{k+1}=\prod_{k=1}^{n} \frac{k+2}{k+1}\prod_{k=1}^{n}\frac{k}{k+1}$$

Then you can easyly conclude ;).

8. Dec 5, 2004

### euclid3.14

Thanks for the help. I've written the sum as (n+2)/ 2(n+1) which tends to 1/2 as n tends to infinity.

I also got the odd term to converge to one as p(2n+1) = p2n*a(2n+1) and as p2n tends to 1 and a(2n+1) tends to 1 aswell then the odd term tends to one!

9. Dec 5, 2004

### euclid3.14

Thanks for the help. I've written the sum as (n+2)/ 2(n+1) which tends to 1/2 as n tends to infinity.

I also got the odd term to converge to one as p(2n+1) = p2n*a(2n+1) and as p2n tends to 1 and a(2n+1) tends to 1 aswell then the odd term tends to one!