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Make a conjecture about a formula for the product: $\left(1−\frac14\right)\cdot\left(1−\frac19\right)\cdot\cdots\cdot\left(1−\frac1{n^2}\right)$ for all natural numbers $n$ with $n \ge 2$. Then, state a theorem about the formula and use mathematical induction to prove your theorem. Any help appreciated!

I don't know what the conjecture is, thats what I need help with. All I found was the result is always between 0 and 1. I don't know if there should be something more to that or not.

Martin Sleziak
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    Did you try computing the first few products of this form? What results did you get? What is your conjecture? –  Apr 16 '18 at 19:53
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    I don't know what the conjecture is, thats what I need help with. All I found was the result is always between 0 and 1. I don't know if there should be something more to that or not. –  Apr 16 '18 at 19:54
  • Yeah, it should be a lot more. Write out the first four terms and see what they are. –  Apr 16 '18 at 19:55
  • Well for the first four terms I get a result of .75, .8889, .9375, and .96 which multiplied together is .6. I don't know what to do from there. –  Apr 16 '18 at 20:00
  • Write these as fractions and simplify. –  Apr 16 '18 at 20:01
  • As fractions it is 3/4 * 8/9 * 15/16 * 24/25 = 3/5 –  Apr 16 '18 at 20:03
  • @HeathLawrence By the way, the sequence of terms you should be looking for is $$\left(1−\frac14\right),,,,\left(1−\frac14\right)\cdot\left(1−\frac19\right),,,,\left(1−\frac14\right)\cdot\left(1−\frac19\right)\cdot\left(1−\frac1{16}\right), \dots$$ Right now, it looks like you are looking at $$\left(1−\frac14\right),,,,\left(1−\frac19\right),,,,\left(1−\frac1{16}\right),\dots$$user296602 suggested to write out that sequence and hope to spot a pattern – John Doe Apr 16 '18 at 20:03
  • Sorry, I am confused. Why is the sequence of terms have (1-1/4) repeated, then in the next sequence it isn't repeated? I though it would just be continuous to whatever n is so (1-1/4) * (1-1/9) and so on until whatever n is. –  Apr 16 '18 at 20:10
  • @HeathLawrence There is not enough space to write it in the comments, so I have posted it in the answers. Let me know if it makes sense, and how you get on – John Doe Apr 16 '18 at 20:14
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    See also Proving $\prod_{i=2}^{i=n} \left(1-\frac{1}{i^2}\right) = \frac{n+1}{2n}$ by induction or Simplifying the product $\prod\limits_{k=2}^n \left(1-\frac1{k^2}\right)$ – Martin Sleziak Apr 16 '18 at 22:40
  • @MartinSleziak I am not sure if this is a duplicate - it seems OP's main problem was forming the induction hypothesis, rather than proving it, no? – John Doe Apr 17 '18 at 01:10
  • @JohnDoe Perhaps. (To be honest the OP did not say much what his problem actually is.) IIRC when I came here the question already had two close votes as off-topic/missing context. (I guess that you chose this reason in your close vote.) From the two possible outcomes (closure as off-topic/closure as duplicate) I thought that duplicate might be more useful choice, since it at least points the OP (and other users seeing this post) to other questions about the same problem. But very likely there is no much difference between the two possibilities. – Martin Sleziak Apr 17 '18 at 01:17
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    @MartinSleziak Yes, that's right, I had chosen off topic. I should have retracted that when OP was responsive in the comments I suppose, but oh well. And yes, I agree - the second link in particular actually has a nice proof that is different to induction. – John Doe Apr 17 '18 at 01:19
  • @HeathLawrence As you can see from the above comments, some users voted to close your question with the close reason missing context - similar as you saw in some of your previous question. It might help if you edit your question a bit. (I have at least included what you mentioned in your first comment.) – Martin Sleziak Apr 17 '18 at 01:45
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    @JohnDoe It seems after your reopen vote (I suppose it was from you judging by the above comment) the question went to reopen votes review, but the end result was to leave the question closed. – Martin Sleziak Apr 17 '18 at 02:05

1 Answers1

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The terms in your sequence are:

$n={2}$: term is $\left(1-\frac1{4}\right)$

$n={3}$: term is $\left(1-\frac1{4}\right)\times\left(1-\frac1{9}\right)$

$n={4}$: term is $\left(1-\frac1{4}\right)\times\left(1-\frac1{9}\right)\times\left(1-\frac1{16}\right)$

$n={5}$: term is $\left(1-\frac1{4}\right)\times\left(1-\frac1{9}\right)\times\left(1-\frac1{16}\right)\times\left(1-\frac1{25}\right)$

$\cdots$

(Edit: writing a proper answer now that OP has figured it out)

The numbers you got, when written as fractions are: $$\left\{\frac34,\,\,\frac23,\,\,\frac58,\,\,\frac35,\cdots\right\}=\left\{\frac34,\,\,\frac46,\,\,\frac58,\,\,\frac6{10},\cdots\right\}$$Here, the numerator clearly has nth term of $n+1$, and the denominator has $2n$ as the nth term. So we conjecture that the general term is $$\frac{n+1}{2n}$$ What is left is to prove this by induction.

John Doe
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  • well I tried taking the result of each n term so for n = 2 its .75, then n=3 it is .667, then n = 4 is .625, n=5 is .6. The pattern I see which I guess is obvious is that the answer continues to get smaller as n gets larger. I don't see any other patterns, which i'm sure my pattern doesn't help at all :( –  Apr 16 '18 at 20:20
  • Ok, that is good. The numbers you got, when written as fractions are: $$\frac34,,,\frac23,,,\frac58,,,\frac35$$Now i'll give you a hint - instead of writing them like that, i will write them like this: $$\frac34,,,\frac46,,,\frac58,,,\frac6{10}$$Do you see what is happening for each term? Do you see a pattern in the numerators? And the denominators? – John Doe Apr 16 '18 at 20:32
  • Wow ok yes I see the pattern in the answers. So for whatever n is, the result will be (n+1) / (2n). So then from there would my conjecture be that if n is a natural number greater than 2, then (1−1/4)⋅(1−1/9)⋅⋯⋅(1−1/n^2) = (n+1) / (2n) and from there just do normal mathematical induction? –  Apr 16 '18 at 20:39
  • @HeathLawrence sorry for the late reply, but yes, thats it! – John Doe Apr 16 '18 at 21:01
  • Thank you so much! –  Apr 16 '18 at 21:04