Geometric Distribution Probability Problem 
Suppose that $X$ is a geometric random variable with parameter (probability of success) $p$.
Show that $\Pr(X > a+b \mid X>a) = \Pr(X>b)$

First I thought I'd start by calculating $\Pr(X>n)$ where $n=a+b$:
$$\Pr(X > n) = p_{n+1} + p_{n+2} + \cdots = ?\tag{1}$$
But I don't know how to determine the limit of equation (1).  I know for an infinite geometric series starting at index zero:
$$\sum\limits_{n=0}^\infty ax^n=\cfrac{a}{1-x}\text{ for }|X|<1$$
But don't know what to do when index starts at $n$.
Next I thought I'd do:
$$\Pr(X > a+b) \mid X > a) = \cfrac{  \Pr[ (X > a+b) \cap (X > a)]  }{\Pr(X > a)}$$
$$=\cfrac{\Pr(X > a+b)}{\Pr(X > a)}$$
and substitute my result from equation (1).  Any help appreciated in advance.  Thank you.
 A: There are two ways to get $\Pr(X\gt n)$, the easy way and the hard way. 
Since there is something to be learned from both, we do it both ways.
The hard way: The probability that $X=n+1$ is the probability of $n$ failures in a row, then success. This is $(1-p)^np$.
Similarly, the probability that $X=n+2$ is the probability of $n+1$ failures in a row, then success. This has probability $(1-p)^{n+1}p$.
Continue, and add up. There is a $(1-p)^n p$ "in" each term. Take it out as a common factor. So we get
$$(1-p)^n p\left(1+(1-p)+(1-p)^2 +\cdots\right).$$
The geometric series inside the brackets has sum $\frac{1}{1-(1-p)}=\frac{1}{p}$.
The easy way: We have $X\gt n$ precisely if we get $n$ failures in a row, probability $(1-p)^n$.
A: Here's how to handle an infinite geometric series when the index starts at $n$ instead of $0$:
\begin{align}
\sum_{k=n}^\infty ax^k & = ax^n + ax^{n+1} + ax^{n+2} + ax^{n+3}+\cdots \\[10pt]
& = (ax^n) + (ax^n)x +(ax^n)x^2 + (ax^n)x^3+\cdots \\[10pt]
& = b + bx + bx^2 + bx^3 + \cdots
\end{align}
Now it starts at index $0$.  And of course $b$ is $ax^n$.
It looks as if you can do the rest.
Second method: You mentioned that the probability of "success" is $p$.  That means the probability of success on each trial is $p$.
If $X$ is defined as the number of trials needed to get one success, then the event $X>n$ is the same as the event of failure on all of the first $n$ trials, so that probability of that is $(1-p)^n$.
If $X$ is defined as the number of trials needed to get one failure, then the event $X>n$ is the same as the event of success on all of the first $n$ trials, so the probability of that is $p^n$.
