The Fibonacci sequence starts out $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, \ldots$
so the given sum is
$$\frac{1}{1} + \frac{3}{1} + \frac{8}{2} + \frac{21}{3} + \frac{55}{5} + \frac{144}{8} + \frac{377}{13} + \frac{987}{21} + \frac{2584}{34} + \frac{6765}{55},$$
which equals $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319$.

The Fibonacci sequence starts out $1, 1, 2, 3, 5, 8, 13, 21, 34, 55$, so the given sum starts out
$$\frac{1}{1} + \frac{3}{1} + \frac{8}{2} + \frac{21}{3} + \frac{55}{5} = 1 + 3 + 4 + 7 + 11.$$

It appears that these summands satisfy the same recurrence relation, namely
$$\frac{F_{2n}}{F_n} = \frac{F_{2(n-1)}}{F_{n-1}} + \frac{F_{2(n-2)}}{F_{n-2}}.$$

To prove the identity for ($L_n$) displayed above, recall Binet's formula, $F_n = \frac{1}{\sqrt{5}}(\phi^n - \psi^n)$, where $\phi = \frac{1+\sqrt{5}}{2}$ and $\psi = \frac{1-\sqrt{5}}{2}$ are the roots of the polynomial $x^2 - x - 1$. Then
$$L_n = \frac{F_{2n}}{F_n} = \frac{\phi^{2n} - \psi^{2n}}{\phi^n - \psi^n} = \phi^n + \psi^n.$$
Therefore
$$$$
$$\begin{align*} L_{n-1} + L_{n-2} &= \phi^{n-1} + \phi^{n-2} + \psi^{n-1} + \psi^{n-2} \\ &= \phi^{n-2}(\phi + 1) + \psi^{n-2}(\psi + 1) \\ &= \phi^{n-2} \cdot \phi^2 + \psi^{n-2} \cdot \psi^2 \\ &= \phi^n + \psi^n = L_n. \end{align*}$$