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$\sum_{t=0}^{\infty}{t\mathcal{\gamma}(1-\gamma)^{tC-1}} = \frac{1}{\gamma C}$ where C is the arbitrary term (constant).

It is the expected time $t (t\ge 0)$ to send $L (L=tC)$ length packet, which the time has geometric distribution possibility.

How It can be possible? I tried to use the way of making convergence of geometric distribution like $\sum_{t=1}^{\infty}{t\mathcal{\gamma}(1-\gamma)^{t-1}} = \frac{1}{\gamma }$.

So my thought is $\gamma\sum_{t=0}^{\infty} {\frac{\partial}{\partial \gamma}}(-1)\frac{1}{C}(1-\gamma)^{tC} = \gamma \bigg(\frac{\partial}{\partial \gamma}(-1)\frac{1}{C}\frac{1}{1-(1-\gamma)^{C}}\bigg)$

I don't think above equation is going to be $\frac{1}{\gamma C}$.

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    It could be $\frac{1}{\gamma C^2}$ (see my updated answer). – 2017-01-26

1 Answers 1

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Hint

$$A=\sum_{t=0}^{\infty}{t\mathcal{\gamma}(1-\gamma)^{tC-1}} =\frac \gamma {1-\gamma}\sum_{t=0}^{\infty}{t(1-\gamma)^{tC}}=\frac \gamma {1-\gamma}\sum_{t=0}^{\infty}{t\left((1-\gamma)^{C}\right)}^t$$ Define $x=(1-\gamma)^{C}$ to make $$A=\frac \gamma {1-\gamma}\sum_{t=0}^{\infty}t x^t=\frac {\gamma x}{1-\gamma}\sum_{t=0}^{\infty}t x^{t-1}=\frac {\gamma x}{1-\gamma}\left(\sum_{t=0}^{\infty} x^{t}\right)'$$

Edit

The end result being $$A=\frac{\gamma (1-\gamma )^{C-1}}{\left(1-(1-\gamma )^C\right)^2}$$ Now, if $\gamma$ is small compared to $1$, a series expansion would give $$A=\frac{1}{\gamma C^2}+\frac{1}{12} \gamma \left(\frac{1}{C^2}-1\right)+\frac{1}{12} \gamma ^2 \left(\frac{1}{C^2}-1\right)+O\left(\gamma ^3\right)$$ which reduced to the first term would be $$A=\frac{1}{\gamma C^2}+O\left(\gamma ^1\right)$$

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    But this is exactly what the OP explains, no? – 2017-01-25