Proving that for a geometric distributed random variable $P(X\gt n+k-1 | X\gt n-1) = P(X\gt k)$ I can reduce the left hand side to $(1-p)^k$. However, on the right hand side I am getting $(1-p)^{k-1}$. In the solution, the book shows the right side follows from the fact that;
$P(X\gt k)$ = $1$ - $P(X\le k)$
            = $1$ - $\sum_{i=0}^k (1-p)^i p$
For a geometric distribution though isn't this incorrect? Shouldn't it be
$1$ - $\sum_{i=0}^{k-1} (1-p)^i p$
Any input would be greatly appreciated!
 A: We will assume that the version of the geometric is the one popular in first courses. So $X$ is the total number of trials up to and including the first success. 
Let us calculate $\Pr(X \gt k)$ without summing a series. We require more than $k$ trials precisely if the first $k$ trials result in failure. The probability of this is $(1-p)^k$. 
The end.
But to respond to your question, the probability of more than $k$ trials is $1$ minus the probability of $\le k$ trials. 
The probability of $\le k$ trials is the probability it takes $1$, plus the probability it takes $2$, and so on up to $k$. If we want to use the summation symbol, 
$$\Pr(X\le k)=\sum_{j=1}^k p(1-p)^{j-1}.$$
If for some reason we want to sum from $0$, we replace $j-1$ by $i$ and end up with
$$\sum_{i=0}^{k-1} p(1-p)^i,$$
precisely as you wrote. 
By the way, the sum is, by the usual formula, equal to $\frac{p-p(1-p)^k}{(1-(1-p)}$, that is, $1-(1-p)^k$. Subtracting from $1$, we get $(1-p)^k$. A lot of trouble for something that we did earlier in one line.
