# Formula for Geometric Progression

Can someone help me understand the idea behind constructing a formula for the following:

For $n\in\mathbb{N}$, $n\geq 2$, find and prove a formula for: $$\prod_{i=2}^n \left(1 - \frac{1}{i^2}\right).$$

Please, Please, I need to know HOW to do this not just how to solve this question.

- Why does I start at 2 and not 1 like regular series, is it because it makes the solution zero?
- Why does $n \geq 2$ have to be true?
- Do you start with $i = 2$ for the first element?

Thanks any help is appreciated!

-
This is utterly mangled. "prove the geometric product" what? What is it you are supposed to prove about the geometric product? As to your other questions: you can start wherever you want; there is no law saying that sums or products have to start with $1$; for your second question, since your statement does not even include an $n$, it's impossible to answer until you post the actual statement you are supposed to establish. – Arturo Magidin Mar 8 '11 at 2:37
@1337holiday: Still incomplete. "prove" what? The expression does not even include an $n$ in it. Right now, it's not even a coherent statement let alone something that can be solved. – Arturo Magidin Mar 8 '11 at 2:40
ma bad should be good now – 1337holiday Mar 8 '11 at 2:42
@1337holiday: No, still no $n$ anywhere. – Arturo Magidin Mar 8 '11 at 2:47
@1337holiday: To your second question: if you take $i=1$, then the first factor would be $0$; finding a formula for a large product, one of whose factors is $0$, would be a rather trivial exercise. – Arturo Magidin Mar 8 '11 at 2:47

You can start a summation or a product anywhere, not just at $1$. There is no law mandating that they must start at $1$.

In this case, if you were to allow $i=1$, then the first factor of the product would be $$\left(1 - \frac{1}{1^2}\right) = 0$$ and so the entire product would be zero, which would make everything rather silly. So you start at $2$ instead (you could instead reindex and consider $$\prod_{j=1}^{n-1}\left(1 - \frac{1}{(j+1)^2}\right)$$ but that just makes the expression look more complicated; better to change the indices instead).

Why do you have $n\geq 2$? Because if $n\lt 2$, then the product has no factors (since no value of $i$ can be at least $2$ and also less than or equal to some $n\lt 2$). The empty product is equal to $1$ by definition, but most people have trouble with that assertion, so why introduce it? Better to stick to $n\geq 2$ so that it is clear there are always factors.

Yes: you start by plugging in $i=2$, then $i=3$, then $i=4$, and so on until you get to $i=n$. For example, with $n=4$, you have \begin{align*} \prod_{i=2}^4\left(1 - \frac{1}{i^2}\right) &= \left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{4^2}\right)\\ &= \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right)\\ &= \left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{15}{16}\right)\\ &= \frac{360}{576} = \frac{5}{8}. \end{align*} As to finding a formula, well, you might try a few values and see if you spot a pattern; then you can try using induction to prove it.

Added. Okay, you say you are stuck, you don't see a pattern. This is just

Consider the factors you are multiplying: $1 - \frac{1}{i^2}$. Writing them as a single fraction, you have $$1 - \frac{1}{i^2} = \frac{i^2 - 1}{i^2} = \frac{(i-1)(i+1)}{i^2}.$$

So, what are you doing at each step? Let's write them out explicitly: $$\left(\frac{(1)(3)}{2^2}\right)\left(\frac{(2)(4)}{3^2}\right)\left(\frac{(3)(5)}{4^2}\right)\cdots\left(\frac{(n-1)(n+2)}{n^2}\right).$$ The denominator is easy: you can just rearrange it, throw in a couple of extra factors of $1$ (who cares if you multiply by $1$? Doesn't do anything) and write: \begin{align*} &(2)(2)(3)(3)(4)(4)\cdots(n)(n)\\ &=(2)(3)(4)\cdots(n)(2)(3)(4)\cdots(n)= \Bigl((1)(2)(3)(4)\cdots(n)\Bigr)\Bigl((1)(2)(3)(4)\cdots(n)\Bigr). \end{align*} Now, what do you call the product of all the positive integers from $1$ through $n$?

Great! We now have a formula for the denominator.

What about the numerator? Again, reorder them putting first all the first factors, and next all the second factors; we have: \begin{align*} &(1)(3)(2)(4)(3)(5)(4)(6)\cdots (n-1)(n+1)\\ &= \Bigl( (1)(2)(3)(4)\cdots(n-1)\Bigr)\Bigl((3)(4)(5)\cdots(n+1)\Bigr). \end{align*} Hmmm... Looks like another set of those "products of all positive integers from $1$ up to something." Now, granted, the second factor isn't that, but just throw in a $(1)(2)$, and then divide by $2$ to keep it honest: $$\Bigl((1)(2)(3)(4)\cdots(n-1)\Bigr)\Bigl((1)(2)(3)(4)(5)\cdots(n+1)\Bigr)\left(\frac{1}{2}\right).$$ Now, express it using the same kind of symbol as the denominator.

Now, step back and look at what you get. Is there some simplification/cancellation that you can make to get something simple? What?

-
This is Exactly where i got to and now im stuck at 5/8 lol. I tried induction but i do not think im doing it right. I did k + (k+1) on the formula that was given but that just does the product for the last 2 (it would multiply 8/9 and 15/16 but not all the way to 3/4). If you can help me out some more that would be cool. – 1337holiday Mar 8 '11 at 2:59
@1337holiday: See DJC's answer. Can you express the denominator and the numerator as some standard expression, say, factorials? – Arturo Magidin Mar 8 '11 at 3:01
No i dont think i can. His answer is a good hint but can you show me how you would solve for n=4 and get 5/8 with his solution? Thats my only barrier to understanding this now, my prof never taught us this and im trying to learn it, thanks. – 1337holiday Mar 8 '11 at 3:07
@1337holiday: No, I can't solve it for n=4 without solving the general problem. – Arturo Magidin Mar 8 '11 at 3:17
Awesome!! But one thing, how is this different from a geometric progression (summation)? I tried using the same principle to come up with the solution but i couldnt. – 1337holiday Mar 8 '11 at 3:17

HINT: $$\prod_{i=2}^n \left( 1 - \frac{1}{i^2}\right) = \prod_{i=2}^n \left(\frac{i^2 - 1}{i^2} \right)=\prod_{i=2}^n \frac{(i-1)(i+1)}{i^2} = \frac{\displaystyle\prod_{i=2}^n (i-1) \prod_{i=2}^n (i+1)}{\left(\displaystyle\prod_{i=2}^n i\right)^2}.$$

-
I cant believe i didnt see that. I have one request, in Arturo Magidin's answer above, he arrived at 5/8. Can you show me how you would arrive at that? This would clear up a lot of things. – 1337holiday Mar 8 '11 at 3:05
You should try this on your own, write down what you get, and then ask for help only if you get stuck on a step. – JavaMan Mar 8 '11 at 3:13
I feel Arturo should get the checkmark here. If there is a way to change it, his answer is MUCH better than my little hint! – JavaMan Mar 8 '11 at 3:19