Probability that the eventually a six on a dice will appear. 
Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $ m$ and $ n$ be relatively prime positive integers such that $ \frac{m}{n}$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $ m+n$.

If you roll a die $r$ times, the probability of getting a six on the $r$th try is:
$\frac{1}{6} \cdot \frac{5^{r-1}}{6^{r-1}}$
Now I believe casework for Dave so:
1 roll: $ \frac{1}{6} $, 2 rolls: $\frac{1}{6} \frac{5}{6}$, 3 rolls: $\frac{1}{6} \frac{5^2}{6^2}$, .... , $r$ rolls, $\frac{1}{6} \frac{5^{r-1}}{6^{r-1}}$
I would add this up, but still, we never found $r$?
HINTS ONLY
EDIT
As pointed by 5xum, $L=1$ isn't possible with $D = L - 1$. So I got:
$P = 1 \cdot (1/6)\sum_{r=2}^{\infty} (5/6)^{r-2} + 1  + ...$
But $P > 1$ already, which is impossible?
 A: Let $D$ denote the random variable telling how many rolls Dave took to roll a $6$, and let $L$ denote the variable telling how many rolls Linda took.
Then what you calculated were $P(D=k)$ for certain values of $k$. For example, the probability that Dave took $5$ rolls to roll a $6$ is
$$\frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac16$$
What you need to calculate is the probability that $D=L\pm1$ or $D=L$, which is the probability
$$P(D=1\wedge (L=1\vee L=2)) + P(D=2\wedge (L=1\vee L=2\vee L=3)) + \dots$$
and the sum is actually infinite. I think writing it down may simplify it a bit, and you will be summing power series, which is not that hard...
A: Let $D, L$ be the obvious random variables.  Then the desired probability is
\begin{align}
P & = P(D = 1, L = 1) + P(D = 1, L = 2) + P(D = 2, L = 1) \\
  & + P(D = 2, L = 2) + P(D = 2, L = 3) + P(D = 3, L = 2) \\
  & + P(D = 3, L = 3) + P(D = 3, L = 4) + P(D = 4, L = 3) \\
  & + \cdots
\end{align}
Each row constitutes an element in a geometric series.  Find the first row sum and the ratio, and the rest is trivial.
A: If $D$ and $L$ denote the number of rolls needed by Dave and Linda and $E$ denotes the event then:$$P(E)=P(D=L)+P(D=L-1)+P(D=L+1)$$
This with $P(D=L-1)=P(D=L+1)$ on base of symmetry, and with:
$$P(D=L)=\sum_{k=1}^{\infty}P(D=k\wedge L=k)=\sum_{k=1}^{\infty}P(D=k)P(L=k)$$For $P(D=L-1)$ you can also find such an expression.

edit:
$$P\left(D=L\right)=\sum_{k=1}^{\infty}P\left(D=k\right)P\left(L=k\right)=\sum_{k=1}^{\infty}\left[\left(\frac{5}{6}\right)^{k-1}\frac{1}{6}\right]^{2}=$$$$\frac{1}{36}\sum_{k=1}^{\infty}\left(\frac{25}{36}\right)^{k-1}=\frac{1}{36}\frac{1}{1-\frac{25}{36}}=\frac{1}{11}$$
As said: likewise you can find $P(D=L-1)$.
