## INdexing in the sum of a geometric series in derivation of geometric dist cdf

Hi, I can't see how the indexing is working in this derivation. The question is: Show that the cdf for a geometric random variable is given by $F_{X}(t) = P(X \leq t) = 1 - (1-p)^{[t]}$, where $[t]$ denotes the greatest integer in $t, t\geq 0$.

The derivation is given as $F_{X}(t) = P(X \leq t) = p \sum_{s=o}^{[t]}(1-p)^{s}$

But $\sum_{s=o}^{[t]}(1-p)^{s} = \frac{1-(1-p)^{[t]}}{1-(1-p)} =\frac{1-(1-p)^{[t]}}{p}$

From which the result follows.

However, the sum of a the first n terms in a geometric series is: $\sum_{k=0}^{n-1}ar^k = a \cdot \frac{1-r^n}{1-r}$

So if we are saying that $s=k, a = 1, r = (1-p)$ and $n-1 = [t]$

Then I can't see how this makes sense since it seems we are summing over $[t]+1$ terms, from $s = 0$ to $s = [t]$

Or in other words, if $n-1 = [t]$, why isn't the closed form of the sum over $s$ as follows: $\frac{1-(1-p)^{[t]+1}}{1-(1-p)}$ ?

Thanks in advance for any insights. MD