How do I show this series converges to 36?

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SUMMARY

The discussion focuses on evaluating the expected value μ = E(x) for a discrete random variable X with the probability density function f(x) = (1/6)(5/6)^(x-1) for x = 1, 2, 3, ... The user attempts to compute the expectation using the formula μ = Ʃ xf(x) but encounters difficulties when trying to sum the series. A solution is provided, suggesting the user shift the summation index and solve for S, ultimately leading to the conclusion that the series converges to 36.

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Homework Statement


I am currently in a calc-based stats class and we are going over probability models and discrete distributions. I ran into this problem that I'm stuck on.

Let X have the p.d.f. f(x) = (1/6)(5/6)^(x-1) ; x = 1, 2, 3,...
Evaluate μ = E(x).

Homework Equations


EXPECTATION of discrete random variable:
μ = E(x) = Ʃ xf(x)
you are summing all possible x ie 1 to infinity in this case

The Attempt at a Solution


I started off by plugging everything into the above formula as follows.
Ʃ (x)(1/6)(5/6)^(x-1) {x, 1, infinity}
then just pulled out the constant...
(1/6) Ʃ (x)(5/6)^(x-1)
that's were I got stuck. I know the series converges to 36 because of wolfram but I don't know how to get to that. I feel like I'm missing something. I know how to solve the geometric series Ʃ (5/6)^(x-1) obviously but not really with the x tacked on...

I tried setting the series equal to S and then taking a derivative of both sides... I played around with it and somehow got S=S...how helpful, must have done something wrong.

Anyway, I'm not sure were to go from here. I'm obviously very rusty on my series and am probably missing something obvious. Any nudges in the right direction would greatly be appreciated.

Thank you very much!
 
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You are on the right track by setting the sum to S. Instead of taking the derivative, try shifting the summation index by one, you get S=5/6*S+residue. Then solve for S.
 

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