Proof by induction for $\left(1−\frac14\right)\cdot\left(1−\frac19\right)\cdots\left(1−\frac1{n^2}\right)$ for all natural numbers $n$ with $n \ge 2$

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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.

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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.