Proof for expected value of geometric RV without using derivates or other "fancy" methods Is it even possible? I'm guessing it is, but I get stuck very early on:
$E[X]=\sum_{k=1}^\infty kp(1-p)^{k-1}=\sum_{k=1}^\infty k(1-q)(q)^{k-1}=\sum_{k=1}^\infty k(q^{k-1}-q^k)=\sum_{k=0}^\infty (k+1)(q^{k}-q^{k+1})$
I'm trying to make stuff disappear using methods similar to what we use in the case of the telescoping series but I can't do it.
All help/hints are appreciated. Thanks:)
 A: You can prove that
$$E(X) = \sum_{k = 1}^{\infty} P(X\geq k)\text{.}$$
This holds for every random variable which has values in $\mathbb{N}_0$!
Then
$$P(X\geq k) = \sum_{j = k}^\infty (1-q)^{j-1}q = (1-q)^{k-1}$$
and
$$E(X) = \sum_{k=1}^\infty (1-q)^{k-1} = \frac{1}{q}\text{.}$$
A: Suppose that $p\ne 0$. We assume without proof that the expectation exists. Note that $X$ is the number of trials until the first success. Let $a=E(X)$.
With probability $p$, we have success on the first trial. The expectation of $X$, given that we had success on the first trial, is $1$.
With probability $1-p$, we have a failure on the first trial. The expectation of $X$, given we had a failure on the first trial, is $1+a$. Thus
$$a=(p)(1)+(1-p)(1+a).$$
This is a linear equation in $a$. Solve. 
Remark: Alternately, we have 
$$E(X)=p\left(1+2q+3q^2+4q^3+5q^4+\cdots\right).$$
We calculate the sum $s$, where 
$$s=1+2q+3q^2+4q^3+5q^4+\cdots.$$
Note that 
$$sq=q+2q^2+3q^3+4q^4+\cdots.$$
Subtract. We get
$$s(1-q)=1+q+q^2+q^3+q^4+\cdots =\frac{1}{1-q}.$$
Thus $s=\frac{1}{(1-q)^2}=\frac{1}{p^2}$,
and therefore $E(X)=p\cdot \frac{1}{p^2}=\frac{1}{p}$.
We can modify this argument by looking at partial sums 
$$1+2q+3q^2+\cdots +nq^n,$$
finding an explicit expression in basically the same way, and taking the limit as $n\to\infty$.  
