Geometric Series with probability

1. Mar 16, 2009

needhelp83

Using the formula for the sum of geometric series, show that the values of p(n) sum to 1

p(n)=$$(1 - \alpha)^n \alpha$$

My attempt:
$$\alpha \sum^\infty_{{\bf n=0}} (1- \alpha)^n$$

I am not sure where to go from here. Any help to show this is true!

2. Mar 16, 2009

yyat

Do you know the formula

$$\sum_{i=0}^{\infty}x^i=\frac{1}{1-x}$$ valid for $$|x|<1$$ ?

Apply it and you are almost done.

3. Mar 16, 2009

Dick

What's the sum of the geometric series r^n? Now just put r=1-alpha.

4. Mar 16, 2009

needhelp83

$$\alpha \sum_{i=0}^{\infty}x^i=\frac{1}{1-(1-\alpha)} -1 = \frac{(1-\alpha)}{1-(1- \alpha)}$$

I think I got it. Does this look right.

5. Mar 16, 2009

Dick

Not really. Can you fix it? Sum 0 to infinity of r^n is 1/(1-r) for |r|<1 as I recall.

6. Mar 17, 2009

needhelp83

$$\alpha \frac{1}{1-(1-\alpha)} = \frac{\alpha}{0 + \alpha}$$

7. Mar 17, 2009

Dick

And alpha/alpha=?

8. Mar 17, 2009

needhelp83

1. Thanks everybody for the help on this one.

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