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For a M/M/1 queue, calculating the estimated number of jobs $n$ in the queue is given by:

$$E[n] = \sum_{i=1}^{\infty} p_i i = \sum_{i=1}^{\infty} \rho^i (1-\rho) i .$$

The final result for a M/M/1 queue is:

$$E[n] = \frac{\rho}{(1-\rho)}.$$

How is it possible to derive this last step from the formulas above?

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    Do you want to a proof of the identity $\sum_{i=1}^{\infty} i \rho^i (1-\rho) = \frac{\rho}{1-\rho}$?2012-01-30
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    lack of dependence on $n$ is a bit confusing2012-01-30
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    The following related posts: [28331](http://math.stackexchange.com/q/28331), [20418](http://math.stackexchange.com/q/20418), [11464](http://math.stackexchange.com/q/11464). They do not answer your exact question but the methods can be modified in a straightforward way. Also [60660](http://math.stackexchange.com/q/60660) is relevant.2012-01-30
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    @Srivatsan: yes, thats what I want.2012-01-30
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    @Il y a: Would you be happier with $E[n] = \sum_{n=1}^{\infty} p_n n = \sum_{n=1}^{\infty} \rho^n (1-\rho) n = \frac{\rho}{(1-\rho)}$?2012-02-01
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    @Ilya: $n$ is a *random variable* denoting the number of jobs in the system (I have seen $N$ used more commonly), and $E[n]$ is the expected number of jobs in the system. The right hand side gives an expression for this expectation in terms of some other parameter governing the system (the "load factor" $\rho$). Naturally, the right hand side shouldn't involve $n$.2012-02-01

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Let $$\begin{align} f(\rho)&=\sum_{i=1}^\infty i \rho^i (1-\rho)\\ &=(1-\rho)\sum_{i=1}^\infty i \rho^i \\ &=(1-\rho)\rho\sum_{i=1}^\infty i \rho^{i-1} \\ &=(1-\rho)\rho\sum_{i=1}^\infty \frac{d}{d\rho} (\rho^i) \\ &=(1-\rho)\rho\frac{d}{d\rho}\left(\sum_{i=1}^\infty \rho^i\right) \\ &=(1-\rho)\rho\frac{d}{d\rho}\left(\frac{\rho}{1-\rho}\right) \\ &=(1-\rho)\rho\frac{1}{(1-\rho)^2} \\ &=\left(\frac{\rho}{1-\rho}\right) \\ \end{align}$$

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To prove $\sum_{i=1}^{\infty} i \rho^i (1-\rho) = \frac{\rho}{1-\rho}$ (which is meaningful at least for the $0 \le \rho \lt 1$ required for this problem):

Let $f( \rho )=\sum_{i=1}^{\infty} i \rho^i (1-\rho)$.

Let $g( \rho )= f( \rho )-\rho f( \rho )$ and so $g( \rho ) = \sum_{i=1}^{\infty} \rho^i (1-\rho)$ as each term almost cancels.

Then $g( \rho )-\rho g( \rho ) = \rho (1-\rho)$ when all the other terms cancel, so $g( \rho ) = \rho$.

So $f( \rho )-\rho f( \rho ) = \rho $, so $f( \rho ) =\frac{\rho}{1-\rho}$.

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    @did I don't know what happened there2012-08-17