I am given the recurrence relation $$F(n) = F(n - 1) + \frac{1}{n}F(n - 2)$$ with $F(0) = \frac{1}{2}$ and $F(1) = \frac{1}{3}$ and asked to solve it. What I have done so far is let $E(n) = n!F(n)$ and multiply by $n!$ on both sides of the equation giving \begin{equation*} \begin{split} n!F(n) &= n!F(n-1) + n!\frac{1}{n}F(n-2)\\ n!F(n) &= n(n-1)!F(n-1) + (n-1)(n-2)!F(n-2)\\ E(n) &= nE(n - 1) + (n-1)E(n - 2)\\ E(n) - E(n - 1) &= (n - 1)\left[E(n - 1) + E(n - 2)\right]\\ \end{split} \end{equation*}
Now, let $G(n) = E(n) - E(n -1)$, then $$G(n) = (n - 1)G(n - 1)$$
From here I have solved $G(n)$ to be $$G(n) = -\frac{(n - 1)!}{6}.$$ But I am having trouble getting back $F(n)$, I am not sure if my initial conditions for $G(n)$ are wrong or what. Any help would be greatly appreciated.