(1-1/2)(1-1/3)(1-1/4). . . (1-1/n)=1/n
How do i start this?
Sorry, I didn't read the title.
Okay, we have the "base" case:
$\displaystyle 1 - \frac{1}{2} = \frac{1}{2}$
So this works for n = 2.
Let's assume this works for some n = k, that is that
$\displaystyle \left ( 1 - \frac{1}{2} \right ) \left ( 1 - \frac{1}{3} \right ) ~ ... ~ \left ( 1 - \frac{1}{k} \right ) = \frac{1}{k}$
Let's see what the n = k + 1 case says:
$\displaystyle \left ( 1 - \frac{1}{2} \right ) \left ( 1 - \frac{1}{3} \right ) ~ ... ~ \left ( 1 - \frac{1}{k} \right ) \left ( 1 - \frac{1}{k + 1} \right )$
By hypothesis, the product of all but the last factor is simply 1/k, so:
$\displaystyle \left ( 1 - \frac{1}{2} \right ) \left ( 1 - \frac{1}{3} \right ) ~ ... ~ \left ( 1 - \frac{1}{k} \right ) \left ( 1 - \frac{1}{k + 1} \right ) = \frac{1}{k} \cdot \left ( 1 - \frac{1}{k + 1} \right )$
Simplifying a bit:
$\displaystyle = \frac{1}{k} \cdot \frac{(k + 1) - 1}{k + 1}$
$\displaystyle = \frac{1}{k} \cdot \frac{k}{k + 1} = \frac{1}{k + 1}$
as required.
So it is true for n = 2, thus it is true for n = 3, 4, ...
-Dan
Say $\displaystyle n=5$ it is easier to do the following trick.
$\displaystyle \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right) \left( 1 - \frac{1}{4} \right)\left( 1 - \frac{1}{5} \right)$
Combine,
$\displaystyle \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} $
Everything cancels to give,
$\displaystyle \frac{1}{5}$.
Now generalize!