Statistics expectation of discrete variable.

[peripatein]

Hi,

Homework Statement

How may I find to what number Ʃ(m=1 to ∞) m/2^m-1 converges?
Further, suppose I know it converges to 4, why would then E(Y), given that P(Y) = 1/2^m-1, be equal to 2 (thus asserted the answer) and not 4?

Homework Equations

The Attempt at a Solution

I am really not sure how to analyse that series. I have managed to determine, through the ratio test, that it indeed converges, but I am not sure how to break it down so that it could be easily determined to what number it converges. I'd truly appreciate some assistance.

 

[tiny-tim]

hi peripatein!

How may I find to what number Ʃ(m=1 to ∞) m/2^m-1 converges?

calculus?

Further, suppose I know it converges to 4, why would then E(Y), given that P(Y) = 1/2^m-1, be equal to 2 (thus asserted the answer) and not 4?

i don't understand …

what is Y ? ​

 

[peripatein]

Y is just a discrete variable for which the probability is P(Y) = 1/2^m-1. But how is E(Y) 2 (based on the answers in the book)?! Why is it not 4?
And I realize calculus would render the evaluation of Ʃ(m=1 to ∞) m/2m-1 possible, yet I am not sure how to break it down. Could you help?

 

[tiny-tim]

hi peripatein!

And I realize calculus would render the evaluation of Ʃ(m=1 to ∞) m/2m-1 possible, yet I am not sure how to break it down. Could you help?

start by writing f(x) = ∑ mx^-(m-1)

Y is just a discrete variable for which the probability is P(Y) = 1/2m-1.

do you mean P(Y=m) ?

 

[haruspex]

P(Y) = 1/2^m-1

Do those probabilities add up to 1?