The Lucas numbers $L_n$ are defined recursively as follows: $L_0=2,L_1=1,L_n=L_{n-1}+L_{n-2}$ for $n\geq2$. Let $r=0.21347\dots$, whose digits form the pattern of the Lucas numbers. When the numbers have multiple digits, they will "overlap," so $r=0.2134830\dots$, [b]not[/b] $0.213471118\dots$. Express $r$ as a rational number $\frac{p}{q}$, where $p$ and $q$ are relatively prime.
To express \( r \) as a rational number, we need to compute the infinite series:
\[ r = \sum_{i=0}^{\infty} \frac{L_i}{10^{i+1}} \]
where \( L_i \) are the Lucas numbers defined by the recurrence relation:
\[ L_0 = 2, \quad L_1 = 1, \quad L_n = L_{n-1} + L_{n-2} \text{ for } n \geq 2. \]

We will generalize this problem by considering a sequence \( X_i \) that satisfies the same recurrence relation:
\[ X_i = X_{i-1} + X_{i-2} \]
and we will find the sum:
\[ Z = \sum_{i=0}^{\infty} \frac{X_i}{n^{i+1}} \]

1. Write out the first few terms of the series:
\[ Z = \frac{X_0}{n} + \frac{X_1}{n^2} + \sum_{i=2}^{\infty} \frac{X_i}{n^{i+1}} \]

2. Using the recurrence relation \( X_i = X_{i-1} + X_{i-2} \), we can express the series as:
\[ Z = \frac{X_0}{n} + \frac{X_1}{n^2} + \sum_{i=2}^{\infty} \frac{X_{i-1} + X_{i-2}}{n^{i+1}} \]

3. Split the sum:
\[ Z = \frac{X_0}{n} + \frac{X_1}{n^2} + \sum_{i=2}^{\infty} \frac{X_{i-1}}{n^{i+1}} + \sum_{i=2}^{\infty} \frac{X_{i-2}}{n^{i+1}} \]

4. Reindex the sums:
\[ Z = \frac{X_0}{n} + \frac{X_1}{n^2} + \sum_{i=1}^{\infty} \frac{X_i}{n^{i+2}} + \sum_{i=0}^{\infty} \frac{X_i}{n^{i+3}} \]

5. Combine the sums:
\[ Z = \frac{X_0}{n} + \frac{X_1}{n^2} + \frac{1}{n} \sum_{i=1}^{\infty} \frac{X_i}{n^{i+1}} + \frac{1}{n^2} \sum_{i=0}^{\infty} \frac{X_i}{n^{i+1}} \]

6. Notice that the sums are similar to \( Z \):
\[ Z = \frac{X_0}{n} + \frac{X_1}{n^2} + \frac{1}{n} (Z - \frac{X_0}{n}) + \frac{1}{n^2} Z \]

7. Simplify the equation:
\[ Z = \frac{X_0}{n} + \frac{X_1}{n^2} + \frac{Z}{n} - \frac{X_0}{n^2} + \frac{Z}{n^2} \]
\[ Z = \frac{X_0}{n} + \frac{X_1}{n^2} - \frac{X_0}{n^2} + \frac{Z}{n} + \frac{Z}{n^2} \]
\[ Z = \frac{X_0}{n} + \frac{X_1 - X_0}{n^2} + \frac{Z}{n} + \frac{Z}{n^2} \]

8. Factor out \( Z \):
\[ Z \left( 1 - \frac{1}{n} - \frac{1}{n^2} \right) = \frac{X_0}{n} + \frac{X_1 - X_0}{n^2} \]

9. Solve for \( Z \):
\[ Z = \frac{\frac{X_0}{n} + \frac{X_1 - X_0}{n^2}}{1 - \frac{1}{n} - \frac{1}{n^2}} \]

10. Plug in \( n = 10 \), \( X_0 = 2 \), and \( X_1 = 1 \):
\[ Z = \frac{\frac{2}{10} + \frac{1 - 2}{10^2}}{1 - \frac{1}{10} - \frac{1}{10^2}} \]
\[ Z = \frac{\frac{2}{10} - \frac{1}{100}}{1 - \frac{1}{10} - \frac{1}{100}} \]
\[ Z = \frac{0.2 - 0.01}{1 - 0.1 - 0.01} \]
\[ Z = \frac{0.19}{0.89} \]
\[ Z = \frac{19}{89} \]

The final answer is \( \boxed{\frac{19}{89}} \)