CDF of $p(1-p)^y$ I need to find the CDF for  
\begin{align}
f(y) = 
\begin{cases}
p(1-p)^y,  & \text{for $y = 0, 1, 2, 3,...$} \\
0, & \text{otherwise}
\end{cases}
\end{align}
Since the sum of $ab^k$ is given as $a{1-b^{m+1}\over 1-b}$, I think that the CDF for this discrete PMF should be 
$p{1-{(1-p)}^\infty\over 1-(1-p)}$
Since $0<p<1$, I think $(1-p)$ should become infinitely smaller until it is effectively zero, leaving me with 
$p{1\over 1-(1-p)}$ 
This would simplify to $F(y)=1/p$ unless I did something wrong- did I? I always get confused working with series and sequences, especially infinite series. As an aside, does anyone have some recommendations for remedial notes on series and sequences?
Thank you. 
 A: $p(1-p)^y~\mathbf 1_{y\in \Bbb N}$ is the probability of obtaining exactly $y$ consecutive 'failures' before the first success in a sequence of iid Bernoulli trials. (AKA a Geometric($0$) distribution)
The Cumulative distribution for this would be the probability of not obtaining $y+1$ consecutive 'failures' before the first success.   Clearly: $1-(1-p)^{y+1}$
You can also find this through the finite Geometric series $$\begin{align}\sum_{k=0}^{\quad y} p(1-p)^k ~=~& \sum_{k=0}^{\quad y} ar^k & \text{where}~ a= 0, r=(1-p)
\\[1ex] =~& \dfrac{a(1-r^{y+1})}{1-r}
\\[1ex] = ~& \dfrac{p~(1-(1-p)^{y+1})}{1-(1-p)}
\\[1ex] = ~& 1-(1-p)^{y+1}\end{align}$$
Thus $$\mathsf P(Y\leq y)  = \begin{cases}1-(1-p)^{y+1} &:& y\in\{0,1,2,...\}
\\ 0 &:& \text{elsewhere}\end{cases}$$
A: It doesn't reduce to $\frac 1p$ but to $1$. There is a $p$ in the numerator as well.
CDF of a function is always $1$ at $\infty$. If you want $F(y)$, sum only until $y$
$\implies f(0)+f(1)+...+f(y) \implies p\frac {1-(1-p)^{y+1}}{1-(1-p)}=1-(1-p)^{y+1}$
