Let $(F_n)_{n\in{N^*}}$ be the Fibonacci sequence defined by
$F_1=1$, $F_2=1$, $F_{n+1}=F_n+F_{n-1}$ for every $n\geq{2}$. Find
the limit: \[ \lim_{n \to \infty}(\sum_{i=1}^n{\frac{F_i}{2^i}}) \]
1. **Define the Fibonacci sequence and the problem:**
   The Fibonacci sequence \((F_n)_{n \in \mathbb{N}^*}\) is defined by:
   \[
   F_1 = 1, \quad F_2 = 1, \quad F_{n+1} = F_n + F_{n-1} \quad \text{for every } n \geq 2.
   \]
   We need to find the limit:
   \[
   \lim_{n \to \infty} \left( \sum_{i=1}^n \frac{F_i}{2^i} \right).
   \]

2. **Express the Fibonacci sequence in its closed form:**
   The closed form of the Fibonacci sequence is given by Binet's formula:
   \[
   F_n = \frac{\phi^n - \psi^n}{\sqrt{5}},
   \]
   where \(\phi = \frac{1 + \sqrt{5}}{2}\) and \(\psi = \frac{1 - \sqrt{5}}{2}\).

3. **Substitute the closed form into the sum:**
   We need to evaluate:
   \[
   \sum_{i=1}^n \frac{F_i}{2^i} = \sum_{i=1}^n \frac{1}{2^i} \cdot \frac{\phi^i - \psi^i}{\sqrt{5}} = \frac{1}{\sqrt{5}} \sum_{i=1}^n \left( \frac{\phi^i}{2^i} - \frac{\psi^i}{2^i} \right).
   \]

4. **Separate the sums:**
   \[
   \sum_{i=1}^n \frac{F_i}{2^i} = \frac{1}{\sqrt{5}} \left( \sum_{i=1}^n \left( \frac{\phi}{2} \right)^i - \sum_{i=1}^n \left( \frac{\psi}{2} \right)^i \right).
   \]

5. **Evaluate the geometric series:**
   Since \(\left| \frac{\phi}{2} \right| < 1\) and \(\left| \frac{\psi}{2} \right| < 1\), we can use the formula for the sum of a geometric series:
   \[
   \sum_{i=1}^\infty \left( \frac{\phi}{2} \right)^i = \frac{\frac{\phi}{2}}{1 - \frac{\phi}{2}} = \frac{\phi/2}{1 - \phi/2},
   \]
   and
   \[
   \sum_{i=1}^\infty \left( \frac{\psi}{2} \right)^i = \frac{\frac{\psi}{2}}{1 - \frac{\psi}{2}} = \frac{\psi/2}{1 - \psi/2}.
   \]

6. **Simplify the expressions:**
   \[
   \frac{\phi/2}{1 - \phi/2} = \frac{\phi}{2 - \phi},
   \]
   and
   \[
   \frac{\psi/2}{1 - \psi/2} = \frac{\psi}{2 - \psi}.
   \]

7. **Combine the results:**
   \[
   \lim_{n \to \infty} \left( \sum_{i=1}^n \frac{F_i}{2^i} \right) = \frac{1}{\sqrt{5}} \left( \frac{\phi}{2 - \phi} - \frac{\psi}{2 - \psi} \right).
   \]

8. **Substitute the values of \(\phi\) and \(\psi\):**
   \[
   \phi = \frac{1 + \sqrt{5}}{2}, \quad \psi = \frac{1 - \sqrt{5}}{2}.
   \]
   Therefore,
   \[
   2 - \phi = 2 - \frac{1 + \sqrt{5}}{2} = \frac{4 - (1 + \sqrt{5})}{2} = \frac{3 - \sqrt{5}}{2},
   \]
   and
   \[
   2 - \psi = 2 - \frac{1 - \sqrt{5}}{2} = \frac{4 - (1 - \sqrt{5})}{2} = \frac{3 + \sqrt{5}}{2}.
   \]

9. **Calculate the fractions:**
   \[
   \frac{\phi}{2 - \phi} = \frac{\frac{1 + \sqrt{5}}{2}}{\frac{3 - \sqrt{5}}{2}} = \frac{1 + \sqrt{5}}{3 - \sqrt{5}} \cdot \frac{3 + \sqrt{5}}{3 + \sqrt{5}} = \frac{(1 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} = \frac{3 + 3\sqrt{5} + \sqrt{5} + 5}{9 - 5} = \frac{8 + 4\sqrt{5}}{4} = 2 + \sqrt{5},
   \]
   and
   \[
   \frac{\psi}{2 - \psi} = \frac{\frac{1 - \sqrt{5}}{2}}{\frac{3 + \sqrt{5}}{2}} = \frac{1 - \sqrt{5}}{3 + \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{(1 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} = \frac{3 - 3\sqrt{5} - \sqrt{5} + 5}{9 - 5} = \frac{8 - 4\sqrt{5}}{4} = 2 - \sqrt{5}.
   \]

10. **Combine the results:**
    \[
    \frac{1}{\sqrt{5}} \left( (2 + \sqrt{5}) - (2 - \sqrt{5}) \right) = \frac{1}{\sqrt{5}} \cdot 2\sqrt{5} = 2.
    \]

The final answer is \(\boxed{2}\).