# Math Help - Help Mathematical Proofs pleasee

1. ## Help Mathematical Proofs pleasee

Heyy Forum
Im having difficulty figuring out this question soo any help would be greatly appriciated
Thanks so much

2. Did you try working out the product for different values of $n$?

3. Originally Posted by roninpro
Did you try working out the product for different values of $n$?
Yes i did but i still couldnt seem to figure it out

4. I computed the values for $2\leq n\leq 20$:

$\frac{3}{4},\frac{2}{3},\frac{5}{8},\frac{3}{5},\f rac{7}{12},\frac{4}{7},\frac{9}{16},\frac{5}{9},\f rac{11}{20},\frac{6}{11},\frac{13}{24},\frac{7}{13 },\frac{15}{28},\frac{8}{15},\frac{17}{32},\frac{9 }{17},\frac{19}{36},\frac{10}{19},\frac{21}{40}$

Do you see a pattern, at least in the denominators?

5. Originally Posted by jess0517
Heyy Forum
Im having difficulty figuring out this question soo any help would be greatly appriciated
Thanks so much

Do some particular cases: $n=2\Longrightarrow 1-\frac{1}{4}=\frac{3}{4}\,,\,\,n=3\Longrightarrow \frac{3}{4}\cdot\frac{8}{9}=\frac{2}{3}$ , $n=4\Longrightarrow \frac{2}{3}\cdot \frac{15}{16}=\frac{5}{8}$ ,

etc., and now separate in odd and even n's...can you see a pretty clear patern for each case?

Tonio

6. Hello, jess0517!

Find a prove a formula for: . $\displaystyle{\prod ^n_{i=2}\left(1 - \frac{1}{i^2}\right) }$

I cranked out the first few partial products:

. . $\begin{array}{cccc}
n & P(n) \\ \hline \\[-4mm]
2 & \frac{3}{4} \\ \\[-4mm]
3 & \frac{4}{6}\\ \\ [-4mm]
4 & \frac{5}{8}\\ \\[-4mm]
5 & \frac{6}{10}\\ \\[-4mm]
6 & \frac{7}{12} \\ \\[-4mm]
7 & \frac{8}{14} \\
\vdots & \vdots \end{array}$

It appears that: . $P(n) \:=\:\dfrac{n+1}{2n}$

I'll let you supply the proof.

7. ## Direct computation.

I assume "find and prove" means that induction is involved.

But anyway you could find the value of $\prod_{i=2}^{n}(1-\frac{1}{i^2})$ by direct manipulation of the terms of the product $(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})$.

8. Notice that $1-\dfrac1{i^2} = \dfrac{i^2-1}{i^2} = \dfrac{(i-1)(i+1)}{i^2}$. Therefore

$\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\ldots\left(1-\dfrac{1}{n^2}\right) = \dfrac{1\cdot3}{2^2} \cdot \dfrac{2\cdot4}{3^2} \cdot \dfrac{3\cdot5}{4^2} \cdots\dfrac{(n-1)(n+1)}{n^2}.$

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 $\frac12$ at the start and $\frac{n+1}n$ at the end.

9. Originally Posted by Soroban
Hello, jess0517!

I cranked out the first few partial products:

. . $\begin{array}{cccc}
n & P(n) \\ \hline \\[-4mm]
2 & \frac{3}{4} \\ \\[-4mm]
3 & \frac{4}{6}\\ \\ [-4mm]
4 & \frac{5}{8}\\ \\[-4mm]
5 & \frac{6}{10}\\ \\[-4mm]
6 & \frac{7}{12} \\ \\[-4mm]
7 & \frac{8}{14} \\
\vdots & \vdots \end{array}$

It appears that: . $P(n) \:=\:\dfrac{n+1}{2n}$

I'll let you supply the proof.

So using induction would this be correct?.. im not to sure how to write it out formaly on this site but this is how i did it:

[(k+1)+1]/2k =[k+2]/2k = 1/2+1/k = [k+2]/2k ?
Obviously adding a formal base step, induction hypothesis and induction step

10. Thread closed due to this member deleting questions after getting help.