Evaluate the sum $ \sum_{i=0}^{n} (-1)^{n-i} \binom{n}{i} f(i)$

I need to evaluate the following sum, which depends on $n \in \mathbb N$ (call it $S(n)$ if you will)

$$ \sum_{i=0}^{n} (-1)^{n-i} \binom{n}{i} f(i)$$

where $f$ defined over $\mathbb N$ is determined by the identity

$$ \sum_{n \geq 0} f(n) \frac{x^n}{n!} = exp \left ( x+\frac{x^2}{2} \right)$$

This is a problem left as an exercise to the reader in Richard Stanley’s “Enumerative Combinatorics”, in the first few pages of Chapter 1, and I assume it should be simple but none of my approaches, including searching for identities involving binomial coefficients, have worked.

Thank you in advance!

Solutions Collecting From Web of "Evaluate the sum $ \sum_{i=0}^{n} (-1)^{n-i} \binom{n}{i} f(i)$"

Hint: $$\left(\displaystyle \sum_{l \geq 0} f(l) \frac{x^l}{l!}\right).\left(\sum_{m \geq 0} (-1)^m \frac{x^m}{m!} \right) = \quad …….$$

Setting
$$
S_n=\sum_{i=0}^n(-1)^{n-i}{n\choose i}f(i),
$$
we have
\begin{eqnarray}
\exp\left(\frac{x^2}{2}\right)&=&\exp\left(x+\frac{x^2}{2}\right)\exp(-x)
=
\left(\sum_{n=0}^\infty\frac{f(n)}{n!}x^n\right)\left(\sum_{n=0}^\infty\frac{(-1)^n}{n!}x^n\right)\\
&=&\sum_{n=0}^\infty\left(\sum_{i=0}^n\frac{f(i)}{i!}\cdot\frac{(-1)^{n-i}}{(n-i)!}\right)x^n
=\sum_{n=0}^\infty\left(\frac{1}{n!}\sum_{i=0}^n(-1)^{n-i}{n\choose i}f(i)\right)x^n\\
&=&\sum_{n=0}^\infty\frac{S_n}{n!}x^n,
\end{eqnarray}
i.e.
$$
\sum_{n=0}^\infty\frac{S_n}{n!}x^n=\exp\left(\frac{x^2}{2}\right)=\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{x^2}{2}\right)^n=\sum_{n=0}^\infty\frac{x^{2n}}{2^nn!}.
$$
It follows that
$$
S_{2n+1}=0,\ S_{2n}=\frac{(2n)!}{2^nn!} \quad \forall n\ge 0.
$$

Here is your $f(n)$

$$ f(n) = n!\sum_{k=0}^{\lfloor n/2 \rfloor}\frac{2^{-k}}{ (n-2k)!\,k! }. $$

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