How do I characterize the distribution of the expected number of periods before the first success in a binomial distribution? I came across this statement in something at work:

An exponential distribution of the length of up time would result from a model in which the probability of failure (down time) is constant through time.

I interpreted this as a binomial distribution where 0 = up and 1 = down, where the probability of a $1$ occurring is $0 < p < 1$. I realize this notation is a little backward, since we don't usually think of "up" as equating with "failure". 
Now I'm trying to characterize the distribution of the length of time that a device is "up". Let $X$ be the discrete probability distribution of the length of the first up time. Then
\begin{align}
Pr(x = 0) &= p \\
Pr(x = 1) &= (1 - p) p \\
Pr(x = 2) &= (1 - p)^2 p \\
\ldots \\
Pr(x = n) &= (1 - p)^{n-1} p \\
\end{align}
Taking the expectation:
\begin{align}
\mathbb{E}[X] &= \displaystyle\sum_{n=0}^{\infty} n (1 - p)^{n-1} p \\
&= p \displaystyle\sum_{n=0}^{\infty} n (1 - p)^{n-1} \\
&= \cfrac{p}{1-p} \displaystyle\sum_{n=0}^{\infty} n (1 - p)^{n} \\
\end{align}
Since
\begin{align}
\displaystyle \lim_{n \to \infty} \left\lvert \cfrac{a_{n+1}}{a_n} \right\rvert
&= \displaystyle \lim_{n \to \infty} \left\lvert \cfrac{(n+1)(1-p)^{n+1}}{n(1-p)^n} \right\rvert \\
&= (1 - p)\displaystyle \lim_{n \to \infty} \left\lvert \cfrac{(n+1)}{n} \right\rvert \\
&= (1 - p)\displaystyle \lim_{n \to \infty} \left\lvert 1 + \cfrac{1}{n} \right\rvert \\
&= 1 - p \\
&< 1
\end{align}
I know the series converges by the Ratio Test. My problem is how to evaluate the infinite series $\displaystyle\sum_{n=0}^{\infty} n (1 - p)^{n}$, given that I know it converges, although I'm also unsure if I interpreted the initial statement correctly.
EDIT: In response to one of the comments about calculating the sum, is this a correct way to do so?
Let $S = \displaystyle \sum_{n=0}^{\infty} n(1 - p)^n$. Then
\begin{align}
S - (1 - p)S &= \left( \displaystyle \sum_{n=0}^{\infty} n(1 - p)^n \right) - \left( (1 - p) \displaystyle \sum_{n=0}^{\infty} n(1 - p)^n \right) \\
pS &= (1 - p) + (1 - p)^2 + (1 - p)^3 + \cdots \\
pS &= \cfrac{1}{1 - (1 - p)} \\
\Rightarrow S &= \cfrac{1}{p^2} \end{align}
 A: An experiment is repeated independently until the first success. Let random variable $X$ be the number of failures before the first success. If the probability of success on any trial is $p$, then for $k=0,1,2,\dots$ we have
$$\Pr(X=k)=p(1-p)^k.$$
It follows that 
$$E(X)=\sum_{k=0}^\infty kp(1-p)^k.\tag{1}$$
So we can evaluate $E(X)$ by finding a sum. Let us do it another way, by conditioning on the result of the first experiment. 
Let $E(X)=a$. The result of the first trial is a success with probability $p$. The conditional expectation of $X$, given that the first trial yields success, is $0$.
With probability $1-p$, the result of the first trial is a failure. In that case, we have wasted a trial, and the expected number of additional trials until the first success is $a$, so the expected total number of trials is $1+a$. By the Law of Total Expectation, we have
$$a=(p)(0)+(1-p)(1+a).$$
Solve for $a$. We get, for $p\ne 0$, that 
$$a=\frac{1-p}{p}=\frac{1}{p}-1.$$
Remark: For fun we can now use the value of $a$ to find the sum $S$ of the series $1+2(1-p)+3(1-p)^2+\cdots$. For by Formula (1) we have
$a=p(1-p)S$, and therefore $S=\frac{a}{p(1-p)}=\frac{1}{p^2}$.
