Solve $f(0)=0,f(1)=1, \forall n\ge2, f(n)= 2f(n-1)+\frac{(-1)^n}{n!}$ using generating functions Use generating functions to find a formula for: $f(0)=0,f(1)=1, \forall n\ge2 ,$we have: $f(n)= 2f(n-1)+\frac{(-1)^n}{n!}$
My attempt:
Let $g(x)$ be the generating function of $f(n)$ after some calculation I get $g(x) = \frac{e^{-x}-1}{1-2x} = (\sum_{n=0}^\infty\frac{(-1)^n}{n!}x^n) (\sum_{n=0}^\infty(2x)^n) - (\sum_{n=0}^\infty(2x)^n) = \sum_{n=0}^\infty((\sum_{k=0}^n\frac{(-1)^k2^{n-k}}{k!})-2^n)x^n$
Is it correct? can i simplify it somehow?
 A: Why use GF at all? It's 
$$
f(n) = f(n-1) + \frac{(-1)^n}{n!2}
$$
Just keep substituting terms and the expression on the right becomes
$$
\frac{1}{2}\sum_{k=2}^{n}\frac{(-1)^k}{k!} = e^{-1} - \sum_{k=n+1}^{\infty}a_k
$$
Where $a_k$ is the same as in the first sum. The remainder term can be upper-bounded
$$
\sum_{k=n+1}^{\infty}a_k \leq \sum_{k=n+1}^{\infty}|a_k| = \frac{1}{(n+1)!} + \frac{1}{(n+1)!(n+2)} + \frac{1}{(n+1)!(n+2)(n+3)} \leq \frac{1}{(n+1)!} \bigg(1 + \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \ldots \bigg)\\ \leq \frac{1}{(n+1)!} \times \sum_{k=0}^{\infty}\bigg(\frac{1}{n+2} \bigg)^k = \frac{1}{(n+1)!} \times \bigg(1+\frac{1}{n+1} \bigg)
$$
The middle step is due to the fact that $(n+1)^2 < (n+1)(n+2)$.
This gives the lower bound on the overall expression. Replacing 1 with -1 in the numerator you can get the upper bound. 
I don't think the orginal sum exists in closed form (partial exponential series).
EDIT: given the corrected problem, you need to divide both sides by $2^n$ and set 
$$
b_n = \frac{a_n}{2^n}
$$
and then substitute each $b_n$, then on RHS you get a sum
$$
\sum_{k=1}^{n}\frac{(-1)^k}{k!2^k} = \sum_{k=1}^{n} \frac{(-0.5)^k}{k!}
$$
Now apply my approach from the above to $e^{-\frac{1}{2}}$. 
