Calculate :

P= (1-1/2)(1-1/3)(1-1/4)........(1-1/2006)(1-1/2007)

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- May 16th 2006, 10:50 PM #1

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- May 17th 2006, 05:28 AM #2

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Originally Posted by**Tenrasmey**

$\displaystyle \left( 1-\frac{1}{2} \right) \left(1-\frac{1}{3} \right)\cdot .... \cdot \left(1-\frac{1}{2007} \right)$

Combine your fractions,

$\displaystyle \left( \frac{1}{2} \right) \left(\frac{2}{3} \right) \left( \frac{3}{4} \right)\cdot ... \cdot \left( \frac{2005}{2006} \right) \left(\frac{2006}{2007} \right)$

Each denominator kills the numerator thus,

$\displaystyle (1)(1)....(1)\left(\frac{1}{2007} \right) =1/2007$

- May 17th 2006, 09:16 AM #3

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Originally Posted by**ThePerfectHacker**

logs it is a telescoping series.

RonL

*telescoping series: A series $\displaystyle \sum_{r=1}^n a_r $ which can be

written in the form:

$\displaystyle

\sum_{r=1}^n a_r =\sum_{r=1}^n (b_{r}-b_{r+1})=b_1-b_{n+1}

$

for some sequence $\displaystyle \{b_r,\ r=1,\dots n+1 \}$

RonL

- May 17th 2006, 01:38 PM #4

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- May 17th 2006, 01:48 PM #5

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Originally Posted by**ThePerfectHacker**

but taking logs is one of the first tricks I would normally employ, after all

it is a quick method of deriving Stirling's approximation to the factorial.

However there must be many cases where this approach will not be

useful.

RonL

- May 17th 2006, 05:04 PM #6