# Proof by Induction?

• Sep 24th 2007, 03:45 AM
WWTL@WHL
Proof by Induction?
Prove that

$\displaystyle {\prod}_{i=2}^n \left(1-\frac{1}{i^2}\right) = \frac{n+1}{2n}$ for integers $\displaystyle n\ge2$

I want to know - what does the $\displaystyle \prod$ mean? Can I do this question through induction? And how does the $\displaystyle i = 2$ affect my working - as opposed to $\displaystyle i=0$ or $\displaystyle i=1$?

I did attempt this, but I haven't got anywhere near close enough. I, sort of, treated the $\displaystyle \prod$ sign as a $\displaystyle \sum$ - so the term to add on to both sides was $\displaystyle \left(1 - \frac{1}{(k+1)^2}\right)$. So no doubt I was completely wrong there.

Help would be much appreciated. Thanks.
• Sep 24th 2007, 04:05 AM
Plato
Sigma, $\displaystyle \sum {}$, stands for SUM where as Pi, $\displaystyle \Pi$ stands for product.
$\displaystyle \prod\limits_{i = 2}^4 {\left( {1 - \frac{1}{{i^2 }}} \right)} = \left( {1 - \frac{1}{{2^2 }}} \right)\left( {1 - \frac{1}{{3^2 }}} \right)\left( {1 - \frac{1}{{4^2 }}} \right)$
• Sep 24th 2007, 05:13 AM
WWTL@WHL
Quote:

Originally Posted by Plato
Sigma, $\displaystyle \sum {}$, stands for SUM where as Pi, $\displaystyle \Pi$ stands for product.
$\displaystyle \prod\limits_{i = 2}^4 {\left( {1 - \frac{1}{{i^2 }}} \right)} = \left( {1 - \frac{1}{{2^2 }}} \right)\left( {1 - \frac{1}{{3^2 }}} \right)\left( {1 - \frac{1}{{4^2 }}} \right)$

Oh ok thanks!

But what's the difference if it was

$\displaystyle \prod\limits_{i = 1}^4 {\left( {1 - \frac{1}{{i^2 }}} \right)}$

i.e. what difference does the i=2 make as opposed to i=1 or i=0 ?

I really should know this :o
• Sep 24th 2007, 05:18 AM
Plato
Firstly, the problem states that the index begins i=2.
If i=1 then the product would equal zero, would it not?
• Sep 24th 2007, 05:51 AM
WWTL@WHL
Quote:

Originally Posted by Plato
Firstly, the problem states that the index begins i=2.

Yes, of course. But I was confused what the i=2 meant.

Quote:

Originally Posted by Plato
If i=1 then the product would equal zero, would it not?

Yes. So under the pi sign, if it's i=3, does that mean that n=3 is the first
term? (Sorry, I know it's probably very simple to understand, but I'm not quite grasping it).

As for the question, I was able to prove it. Thanks. :)

__________________________________________________ _______________

$\displaystyle P_2$

$\displaystyle \prod\limits_{i = 2}^2 {\left( {1 - \frac{1}{4}} \right)} = \frac{3}{4}$

Therefore true for $\displaystyle P_2$, assume it's true for $\displaystyle P_k$

$\displaystyle P_{k+1}$

$\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k+1}{2k} \left(1-\frac{1}{(k+1)^2}\right)$

$\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k+1}{2k} - \frac{k+1}{2k(k+1)^2}$

$\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{(k+1)^3-(k+1)}{2k(k+1)^2}$

$\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k^3+3k^2+2k}{2k(k+1)^2}$

$\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k(k+1)(k+2)}{2k(k+1)^2}$

$\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k+2}{2(k+1)}$

If $\displaystyle P_k$ is true, then $\displaystyle P_{k+1}$ is also true. But since $\displaystyle P_2$ is true, $\displaystyle \rightarrow$ $\displaystyle P_3$ is true, $\displaystyle \rightarrow$ $\displaystyle P_4$ is true...etc. Hence proved.
• Sep 24th 2007, 06:40 AM
DivideBy0
You can't have k as both the upper bound and the index.

It should be

$\displaystyle \prod_{i=2}^{k+1} \left(1-\frac{1}{i^2} \right)=\frac{(k+1)+1}{2(k+1)}$

with $\displaystyle i$ as the index.
• Sep 24th 2007, 07:57 AM
WWTL@WHL
Quite right. Rookie mistake :o

Thanks.