 Solution

Singularities. The complex exponential factors as $F(z)=e^{z/2}{e}^{1/(2z)}$. The second factor has an essential singularity at zero, and multiplication by the nonvanishing holomorphic function $e^{z/2}$ does not remove it. There are no other finite singularities. On the Riemann sphere, infinity is also an essential singularity, since $F(1/\zeta )=F(\zeta )$.

Symmetry and integral formula. The identity $F(1/z)=F(z)$ and uniqueness of the Laurent series on $0<|z|<\infty$ give $F_{-k}=F_k$ for every integer $k$. The Laurent coefficient formula, on the positively oriented unit circle, givesThe imaginary part vanishes by the substitution $\theta \mapsto 2\pi -\theta$. Hence

Polynomial contour integral. For an integer $n\ge 0$ and integer $p$, the binomial theorem gives $(1+z^2{)}^n={\sum }_{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})z^{2j}$. The residue theorem selects $2j-n-p-1=-1$, that is $j=(n+p)/2$. Therefore the correct complete statement isThe printed nonnegative-integer condition alone misses $j\le n$. For example $n=0,p=2$ satisfies that condition, but the integral is zero and the displayed factorial $(-1)!$ is undefined.

Explicit coefficients. Both Taylor series inconverge absolutely and uniformly on every compact annulus away from zero, so their product can be regrouped. For $k\ge 0$, put $r=s+k$ and $s=m$. For arbitrary integer $k$, use the already proved symmetry:Here $I_j$ is a modified Bessel function. The printed series is directly meaningful for $k\ge 0$; for negative $k$ the absolute values, or an appropriately shifted lower summation limit, are essential.