Deriving an Infinite Series: P_e = 5/3

[wildman]

Homework Statement

I am wondering if someone could give me some insight on how the following infinite series was derived:

P_e = \sum_{-\infty}^\infty (1/2)^{2|n|} = -1 + 2 \sum_{n=0}^\infty (1/2)^{2n} = 5/3

Homework Equations

See above

The Attempt at a Solution

I think the -1 comes when n = 0 and the 2 before the sum is because the absolute value of n makes the result symetrical around 0. That is why one can make the sum from 0 to infinity and multiply by 2. Right??
The second sumation must be equal to 4/3. Right? I guess my real question then is how do you find the closed form of this infinite series?

 

[Avodyne]

I think the -1 comes when n = 0 and the 2 before the sum is because the absolute value of n makes the result symetrical around 0. That is why one can make the sum from 0 to infinity and multiply by 2. Right??

Right!

The second sumation must be equal to 4/3. Right?

Right!

I guess my real question then is how do you find the closed form of this infinite series?

Can you do this one?

\sum_{n=0}^\infty x^n

 

[HallsofIvy]

Actually, the "-1" comes from the fact that when n= 0, (1/2)^{2|n|} is equal to 1. Since you are multiplying the sum, from 0 to infinity, by 2, you are getting that twice and need to subract off one.

 

[wildman]

Can you do this one?
\sum_{n=0}^\infty x^n

Yes, that is equal to 1/(1- 1/2) or 2

 

[Avodyne]

It is if x=1/2. What is it for general x? Then, what value of x applies to your problem? (Hint: it's not 1/2.)

 

[wildman]

general x is x^2 and my x is 1/4 so 1/(1-1/4) is 4/3. All right! Thanks!