1. ## Proof By Induction

(1-1/2)(1-1/3)(1-1/4). . . (1-1/n)=1/n

How do i start this?

2. Originally Posted by Edbaseball17
(1-1/2)(1-1/3)(1-1/4). . . (1-1/n)=1/n

How do i start this?
The k-1th term is
$1 - \frac{1}{k} = \frac{k -1}{k}$

So this may be represented in "Pi" notation as:
$\prod_{k = 2}^n \frac{k - 1}{k}$
(I shifted the variable a bit to make the form simpler to look at.)

Can you finish from here?

-Dan

3. so far i have this

$
1 - \frac{1}{2} *1 - \frac{1}{3} *1 - \frac{1}{4}...1 - \frac{1}{k} = \frac{1}{k}
$

$
1 - \frac{1}{2}*1 - \frac{1}{3}*1 - \frac{1}{4}...1 - \frac{1}{k}*1 - \frac{1}{k+1}= \frac{1}{k}*\frac{1}{k+1}
$

4. Originally Posted by Edbaseball17
so far i have this

$
1 - \frac{1}{2} *1 - \frac{1}{3} *1 - \frac{1}{4}...1 - \frac{1}{k} = \frac{1}{k}
$

$
1 - \frac{1}{2}*1 - \frac{1}{3}*1 - \frac{1}{4}...1 - \frac{1}{k}*1 - \frac{1}{k+1}= \frac{1}{k}*\frac{1}{k+1}
$
Sorry, I didn't read the title.

Okay, we have the "base" case:
$1 - \frac{1}{2} = \frac{1}{2}$

So this works for n = 2.

Let's assume this works for some n = k, that is that
$\left ( 1 - \frac{1}{2} \right ) \left ( 1 - \frac{1}{3} \right ) ~ ... ~ \left ( 1 - \frac{1}{k} \right ) = \frac{1}{k}$

Let's see what the n = k + 1 case says:
$\left ( 1 - \frac{1}{2} \right ) \left ( 1 - \frac{1}{3} \right ) ~ ... ~ \left ( 1 - \frac{1}{k} \right ) \left ( 1 - \frac{1}{k + 1} \right )$

By hypothesis, the product of all but the last factor is simply 1/k, so:
$\left ( 1 - \frac{1}{2} \right ) \left ( 1 - \frac{1}{3} \right ) ~ ... ~ \left ( 1 - \frac{1}{k} \right ) \left ( 1 - \frac{1}{k + 1} \right ) = \frac{1}{k} \cdot \left ( 1 - \frac{1}{k + 1} \right )$

Simplifying a bit:
$= \frac{1}{k} \cdot \frac{(k + 1) - 1}{k + 1}$

$= \frac{1}{k} \cdot \frac{k}{k + 1} = \frac{1}{k + 1}$
as required.

So it is true for n = 2, thus it is true for n = 3, 4, ...

-Dan

5. Say $n=5$ it is easier to do the following trick.

$\left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right) \left( 1 - \frac{1}{4} \right)\left( 1 - \frac{1}{5} \right)$
Combine,
$\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5}$
Everything cancels to give,
$\frac{1}{5}$.

Now generalize!