Expected Value of Geometric Random Variable 

Question: Find the expected time to transmit the packet.

If $T$ is the random variable describing the time it takes for transmitting the 
packet, then $$T = \frac{L}{C}$$. And the probability that it takes t seconds to transmit the packet is $$P(T=t)=P(L= Ct) = \mu(1-\mu)^{Ct-1}$$
So I try this formula to get the expected time to transmit the packet.
$$\sum_{t=0}^{\infty}tP(T=t) = \sum_{t=0}^{\infty}t\mu(1-\mu)^{Ct-1} = \frac{-\mu(1-\mu)^{C}}{(\mu-1)((1-\mu)^{C}-1)^2}$$
But the answer is $$E[T]=\frac{1}{\mu C}$$
What is wrong with my equation?
 A: Your primary error lies in the support of $T$.
Because $T=L/C$ and $L\in \{1,2,3,4,,\ldots\}$, therefore the support of $T$ is not integer.   $T\in\{\frac 1C, \frac 2C, \frac 3C, \ldots\}$
To ensure you are summing over integer intervals, use a change of variables.
Then you should have $$\begin{align}\mathsf E(T) ~&=~ {\sum}_{t\in\{\frac 1C, \frac 2C, \ldots\}} t~\mathsf P(T=t) \\[1ex] &=~ \sum_{tC=1}^\infty t \mu\, (1-\mu)^{tC-1} \\[1ex] &=~ C^{-1}\mu(1-\mu)^{-1}\sum_{k=1}^\infty k(1-\mu)^{k}\end{align}$$
Which simplifies, unsurprisingly, to: $~1/C\mu$
A: great question.  so here is how to solve this.
\begin{equation}
\begin{split}
ET= \sum_{l=0}^{\infty}\frac{l}{c}\mu(1-\mu)^{l-1} \\
&=\frac{\mu}{c}\sum_{l=0}^{\infty}l\cdot (1-\mu)^{l-1} \\
&=\frac{\mu}{c}\sum_{l=0}^{\infty}-\frac{d}{d\mu}(1-\mu)^{l}\\
&=\frac{-\mu}{c}\frac{d}{d\mu}\sum_{l=0}^{\infty}(1-\mu)^{l} \\
&=\frac{-\mu}{c}\frac{d}{d\mu}\frac{1}{1-(1-\mu)} \\
&=\frac{\mu}{c}\frac{-d}{d\mu}\frac{1}{\mu}\\
&=\frac{1}{\mu C}
\end{split}
\end{equation}
we can interchange the infinite series and the derivative because we know that $\sum_{0}^{\infty}l(1-\mu)^{l}$ converges uniformly
