Joint probability of geometric random variables 
$X$ and $Y$ are geometric, independent random variables with parameters $x$ and $y$ respectively. Use their joint P.M.F. to compute an approximate summation for $P(X<Y)$.

I've setup the problem: $\sum_{i=1}^j \sum_{j=j}^\infty x(1-x)^{i-1} y(y-1)^{j-1}$
But I have no idea how to simplify further from here. 
 A: For a geometric random variable $Y$ with parameter $y$, 
$$\begin{align}
P\{Y > n\} &= \sum_{i=n+1}^\infty P\{Y = i\} = \sum_{i=n+1}^\infty y(1-y)^{i-1}\\
&= y(1-y)^n[1 + (1-y) + (1-y)^2 + \cdots ]\\
&= y(1-y)^n \times \frac{1}{1-(1-y)}\\
&=(1-y)^n.
\end{align}$$
Actually, the end result is more easily derived by arguing that the event
$\{Y > n\}$ occurs if and only if the first $n$ trials ended in failure,
and the probability of this is just $(1-y)^n$.
Regardless of which way you computed $P\{Y > n\}$, you can compute 
$$P\{X = n, Y> n\} = P\{X = n\}P\{Y > n\} = x(1-x)^{n-1}(1-y)^n$$ and
hence
$$\begin{align}
P\{X < Y\} &= \sum_{n=1}^\infty P\{X = n, Y > n\}\\
&= \sum_{n=1}^\infty x(1-x)^{n-1}(1-y)^n\\
&= x(1-y) [1 + (1-x)(1-y) + ((1-x)(1-y))^2 + \cdots\\
&= x(1-y)\times \frac{1}{1 - (1-x)(1-y)}\\
&= \frac{x(1-y)}{x+y-xy}.
\end{align}$$
This answer too has a nice interpretation. Let us carry out the
$n$-th trials simultaneously. Then, we can ignore all instances
in which failures were recorded on both trials -- the next pair of
trials must be conducted -- and concentrate on the very first time
that success occurred on at least one of the paired trials, possibly
on both trials.  Conditioned on at least one success, an event
of probability $x+y-xy$, (remember $P(A\cup B) = P(A)+P(B)-P(A)P(B)$
for independent events?), the probability that $X < Y$ is just the
probability that $X$ occurred and $Y$ did not, which is $x(1-y)$. Thus
we have
$$P\{X < Y\} = \frac{x(1-y)}{x+y-xy}, ~~ P\{X > Y\} = \frac{(1-x)y}{x+y-xy}, ~~
P\{X = Y\} = \frac{xy}{x+y-xy}$$
without any need for summing series or indeed without even writing down
the joint pmf etc.
