I remember seeing this as a trick before. I tried to recreate it but it's proving to be difficult.
I have a deck of $13$ cards, consisting of ace through king. I have a card number in my head (ace, $2$-$10$, jack, queen, king). What is the probability of finding it in $n$ draws? After each draw I put the card back and shuffle the deck.
As it is essentially drawing with replacement, we notice that any particular draw will be the card we are looking for with probability $\frac{1}{13}$ and will not be with probability $\frac{12}{13}$.
As each draw is independent of one another, the probability of a specific sequence of draws will be the product of the probabilities of each draw individually. E.g. Ace-Ace-Nonace in that order will occur with probability $\frac{1}{13}\cdot\frac{1}{13}\cdot\frac{12}{13}$.
There are two possible interpretations of the question I can think of.
This follows the geometric distribution with $p=\frac{1}{13}$. This corresponds to getting $n-1$ non-ace cards in a row followed by an ace. This occurs then with probability $\frac{12}{13}\cdot\frac{12}{13}\cdots\frac{12}{13}\cdot\frac{1}{13}=(\frac{12}{13})^{n-1}\cdot\frac{1}{13}$
Getting at least one ace within the first $n$ draws means that the first ace we see occurs within the first $n$ draws, so we may simply sum over the geometric distribution up to $n$. $\sum\limits_{k=1}^n(\frac{12}{13})^{k-1}\frac{1}{13}$ which may be simplified using what you know about geometric series.
An alternate method to calculate this is by recognizing that getting at least one ace within the first $n$ draws is the opposite event of getting no aces within the first $n$ draws. Getting no aces corresponds to not getting an ace for the first draw followed by not getting an ace for the second draw followed by...etc...
Getting no aces in $n$ draws has probability then $\frac{12}{13}\cdot\frac{12}{13}\cdots\frac{12}{13}=(\frac{12}{13})^n$ so getting at least one ace in $n$ draws has probability $1-(\frac{12}{13})^n$. Had you simplified your geometric sum above, it would have arrived at this answer as well.
A word of warning, I mentioned that $Pr(A\cap B)=Pr(A)\cdot Pr(B)$, that is the probability event $A$ and event $B$ simultaneously occur is the probability of the product of their respective probabilities. This is true only for independent events, and is in general not true for arbitrary events.