Math Help Forum - View Single Post

hatsoff · #9 April 14th, 2008, 10:41 AM

Quote:

Originally Posted by CaptainBlack

And of course we have:

$\bar{N}=\sum_{r=1}^{\infty} r~ (1-p)^{r-1}p = \frac{1}{p}=n$

Which we could deduce from more general principles but its nice to see the series summed anyway.

RonL

Thank you. This helps a great deal. I see clearly now that I was mistaken, and I see where and why. However, though I do not doubt your conclusion, neither do I understand how you reached it. I therefore have a couple more questions...

Firstly, how did you get from here:

$\sum_{r=1}^{\infty} r~ (1-p)^{r-1}p$

to here:

$\frac{1}{p}$

? Also, would you be able to explain in newbie terms how you got this initial formula:

$\bar{N}=\sum_{r=1}^{\infty} r~ p(r)$

?

Thanks for all your help so far!