Let $a_1,a_2,\ldots$ be a sequence defined by $a_1=a_2=1$ and $a_{n+2}=a_{n+1}+a_n$ for $n\geq 1$.  Find \[\sum_{n=1}^\infty \dfrac{a_n}{4^{n+1}}.\]
1. We start by noting that the sequence \(a_n\) is defined by the recurrence relation \(a_{n+2} = a_{n+1} + a_n\) with initial conditions \(a_1 = 1\) and \(a_2 = 1\). This sequence is known as the Fibonacci sequence, where each term is the sum of the two preceding ones.

2. We need to find the sum \(\sum_{n=1}^\infty \frac{a_n}{4^{n+1}}\). To do this, we use the generating function for the Fibonacci sequence. The generating function for the Fibonacci sequence \(F(x)\) is given by:
   \[
   F(x) = \sum_{n=1}^\infty a_n x^n = \frac{x}{1 - x - x^2}
   \]
   This generating function can be derived from the recurrence relation and initial conditions of the Fibonacci sequence.

3. We substitute \(x = \frac{1}{4}\) into the generating function:
   \[
   F\left(\frac{1}{4}\right) = \frac{\frac{1}{4}}{1 - \frac{1}{4} - \left(\frac{1}{4}\right)^2}
   \]

4. Simplify the denominator:
   \[
   1 - \frac{1}{4} - \frac{1}{16} = \frac{16}{16} - \frac{4}{16} - \frac{1}{16} = \frac{11}{16}
   \]

5. Substitute back into the generating function:
   \[
   F\left(\frac{1}{4}\right) = \frac{\frac{1}{4}}{\frac{11}{16}} = \frac{1}{4} \cdot \frac{16}{11} = \frac{4}{11}
   \]

6. The sum we are interested in is \(\sum_{n=1}^\infty \frac{a_n}{4^{n+1}}\). Notice that this is \(\frac{1}{4}\) times the generating function evaluated at \(\frac{1}{4}\):
   \[
   \sum_{n=1}^\infty \frac{a_n}{4^{n+1}} = \frac{1}{4} \cdot F\left(\frac{1}{4}\right) = \frac{1}{4} \cdot \frac{4}{11} = \frac{1}{11}
   \]

The final answer is \(\boxed{\frac{1}{11}}\).