I computed the values for $\displaystyle 2\leq n\leq 20$:
$\displaystyle \frac{3}{4},\frac{2}{3},\frac{5}{8},\frac{3}{5},\f rac{7}{12},\frac{4}{7},\frac{9}{16},\frac{5}{9},\f rac{11}{20},\frac{6}{11},\frac{13}{24},\frac{7}{13 },\frac{15}{28},\frac{8}{15},\frac{17}{32},\frac{9 }{17},\frac{19}{36},\frac{10}{19},\frac{21}{40}$
Do you see a pattern, at least in the denominators?
Do some particular cases: $\displaystyle n=2\Longrightarrow 1-\frac{1}{4}=\frac{3}{4}\,,\,\,n=3\Longrightarrow \frac{3}{4}\cdot\frac{8}{9}=\frac{2}{3}$ , $\displaystyle n=4\Longrightarrow \frac{2}{3}\cdot \frac{15}{16}=\frac{5}{8}$ ,
etc., and now separate in odd and even n's...can you see a pretty clear patern for each case?
Tonio
Hello, jess0517!
Find a prove a formula for: . $\displaystyle \displaystyle{\prod ^n_{i=2}\left(1 - \frac{1}{i^2}\right) }$
I cranked out the first few partial products:
. . $\displaystyle \begin{array}{cccc}
n & P(n) \\ \hline \\[-4mm]
2 & \frac{3}{4} \\ \\[-4mm]
3 & \frac{4}{6}\\ \\ [-4mm]
4 & \frac{5}{8}\\ \\[-4mm]
5 & \frac{6}{10}\\ \\[-4mm]
6 & \frac{7}{12} \\ \\[-4mm]
7 & \frac{8}{14} \\
\vdots & \vdots \end{array}$
It appears that: .$\displaystyle P(n) \:=\:\dfrac{n+1}{2n}$
I'll let you supply the proof.
I assume "find and prove" means that induction is involved.
But anyway you could find the value of $\displaystyle \prod_{i=2}^{n}(1-\frac{1}{i^2})$ by direct manipulation of the terms of the product $\displaystyle (1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})$.
Notice that $\displaystyle 1-\dfrac1{i^2} = \dfrac{i^2-1}{i^2} = \dfrac{(i-1)(i+1)}{i^2}$. Therefore
$\displaystyle \left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\ldots\left(1-\dfrac{1}{n^2}\right) = \dfrac{1\cdot3}{2^2} \cdot \dfrac{2\cdot4}{3^2} \cdot \dfrac{3\cdot5}{4^2} \cdots\dfrac{(n-1)(n+1)}{n^2}.$
Now start cancelling: each of the two factors in the numerators cancels with a factor in a neighbouring denominator. The only terms left uncancelled are $\displaystyle \frac12$ at the start and $\displaystyle \frac{n+1}n$ at the end.