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For all $n>=2$, the formula $(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})\dots(1-\frac{1}{n^2}) = \frac{n+1}{2n}$

Proof:

Base case: $n = 2$.

$(1-\frac{1}{4}) = 0.75 = \frac{(2)+1}{2(2)}$

The claim holds for $n = 2.$

Inductive step: $n \geq 2$.

Suppose that $(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})\dots(1-\frac{1}{n^2}) = \frac{n+1}{2n}$ for $n \geq 2$

I want to show that $(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})\dots(1-\frac{1}{(n+1)^2}) = \frac{(n+1)+1}{2(n+1)}$

I've gotten it to the point of what I want to show but I don't really know how to do the math to prove this. Also, is there a better way to format fractions on this site? Because I find this a little difficult to read. Thanks for any help!

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  • $\begingroup$ Putting $ around math expressions typically already helps a good bit .. just did that. $\endgroup$ – Bram28 Dec 14 '17 at 19:52
  • $\begingroup$ Try using something like $\frac{a}{b}$ $\endgroup$ – Dapianoman Dec 14 '17 at 19:52
  • $\begingroup$ Some other great commands: \ge $\ge$, \le $\le$, \neq $\neq$. You can find the whole tutorial here. $\endgroup$ – Chase Ryan Taylor Dec 14 '17 at 20:09
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Suppose that $$(1-(1/4))(1-(1/9))(1-(1/16))....(1-(1/n^2)) = (n+1)/(2n).$$

Now just write down the expression on the left hand side for $n+1$:

$$(1-(1/4))(1-(1/9))(1-(1/16))....(1-(1/n^2))\times(1-(1/(n+1)^2))$$

By the inductive assumption you can replace everything to the left of the $\times$ sign with $(n+1)/(2n)$. Now simplify

$$(\frac{n+1}{2n})\times(1-\frac{1}{(n+1)^2})$$

and show that it equals $(n+2)/(2(n+1))$.

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Because $$\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)\dots\left(1-\frac{1}{n^2}\right) =$$ $$=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(n-1)(n+1)}{2^23^24^2...n^2}=$$ $$=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot(n-1)(n+1)}{2^23^24^2\cdot...\cdot n^2}= \frac{n+1}{2n}$$

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If you have $a_2 \circ ..... \circ a_n = f(n)$ and you want to prove that means

$a_2 \circ ..... \circ a_n \circ a_{n+1} = f(n+1)$. That means you want to prove

$(a_2 \circ ..... \circ a_n) \circ a_{n+1} = f(n) \circ a_{n+1} = f(n+1)$.

That can always be a strategy.

So you need to prove

$\frac {n+1}{2n}* (1- \frac 1{(n+1)^2}) = \frac {(n+1)+1}{2(n+1)}=\frac {n+2}{2(n+1)}$.

Can you do that?

Well, just try:

$\frac {n+1}{2n}* (1- \frac 1{(n+1)^2})=$

$\frac {n+1}{2n} - \frac {n+1}{2n(n+1)^2}=$

$\frac {(n+1)^2}{2n(n+1)} - \frac {1}{2n(n+1)}=$

$\frac {(n+1)^2 - 1}{2n(n+1)} =\frac {n^2 + 2n}{2n(n+1)}=\frac {n+2}{2(n+1)}$

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