• July 12th 2010, 01:04 PM
jess0517
Heyy Forum
Im having difficulty figuring out this question soo any help would be greatly appriciated
Thanks so much (Happy)

Attachment 18180
• July 12th 2010, 02:05 PM
roninpro
Did you try working out the product for different values of $n$?
• July 12th 2010, 02:07 PM
jess0517
Quote:

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
• July 12th 2010, 02:08 PM
roninpro
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?
• July 12th 2010, 02:10 PM
tonio
Quote:

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

Attachment 18180

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
• July 12th 2010, 03:57 PM
Soroban
Hello, jess0517!

Quote:

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.

• July 12th 2010, 04:12 PM
melese
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 $(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})$.
• July 13th 2010, 01:21 AM
Opalg
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.
• July 13th 2010, 11:03 AM
jess0517
Quote:

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
• July 14th 2010, 03:26 AM
mr fantastic
Thread closed due to this member deleting questions after getting help.