Geometric Series with probability

[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:
<br /> \alpha<br /> \sum^\infty_{{\bf n=0}}<br /> (1- \alpha)^n<br />

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

 

[yyat]

Do you know the formula

\sum_{i=0}^{\infty}x^i=\frac{1}{1-x} valid for |x|&lt;1 ?

Apply it and you are almost done.

 

[Dick]

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

 

[needhelp83]

<br /> \alpha \sum_{i=0}^{\infty}x^i=\frac{1}{1-(1-\alpha)} -1 = \frac{(1-\alpha)}{1-(1- \alpha)}<br />

I think I got it. Does this look right.

 

[Dick]

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

 

[needhelp83]

<br /> \alpha \frac{1}{1-(1-\alpha)} = \frac{\alpha}{0 + \alpha}<br /> <br />

 

[Dick]

And alpha/alpha=?

 

[needhelp83]

1. Thanks everybody for the help on this one.