Laurent series of $e^{z+1/z}$ What is the Laurent series  of $e^{z+1/z}$?
I had used $$a_k= \frac{1}{2\pi i}\int_c \frac{f(z)}{z^{k+1}}\,dz $$
for a curve $c$ in which we can use $e^z$ as an analytic func. and expanded the $e^{1/z}$ series expansion.
 A: This is an alternative approach yielding an integral formula.
If $$f(z)=e^{z+1/z}=\sum_{n=-\infty}^{\infty} a_nz^n$$ and we take $z=e^{i\theta},$ then you get $$f(e^{i\theta})=e^{2\cos\theta}=\sum_{n=-\infty}^{\infty} a_ne^{in\theta}.$$
So the $a_n$ are the Fourier coefficients of the function $f(\theta)=e^{2\cos(\theta)}.$
So $$a_n=\frac1{2\pi}\int_{-\pi}^{\pi} e^{2\cos(\theta)-in\theta}d\theta$$
You can use $a_n=a_{-n}$ to also see:
$$a_n=\frac1{2\pi}\int_{-\pi}^{\pi} e^{2\cos(\theta)}\cos(n\theta)\,d\theta$$
That might be enough of a closed form for some.
Of course, there are a lot easier ways to get that.

This can also be written in terms of a complex integral, using the residue theorem, with $\gamma(t)=e^{it},$
$$a_n=\frac1{2\pi i}\int_{\gamma} \frac{e^{z+z^{-1}}}{z^{n+1}}\,dz=\frac1{2\pi i}\int_0^{2\pi} e^{2\cos t}e^{-i (n+1)t}\cdot ie^{it}\,dt$$
Yielding the same integral.

This approach assumes the Laurent series converges on $|z|=1.$ That is not hard to prove here.

We can get back the other answers by expanding $e^{2\cos\theta}:$ $$a_n=\frac1{2\pi}\sum_{k=0}^{\infty}\frac{2^k}{k!}\int_{-\pi}^{\pi}\cos^k\theta e^{-in\theta}\,d\theta$$
But: using $\cos(\theta)=\frac{1}2\left(e^{i\theta}+e^{-i\theta}\right)$ and expanding with the binomial theorem, we get: $$\int_{-\pi}^{\pi}\cos^k\theta e^{-in\theta}\,d\theta=\begin{cases}0&|n|>k\\0&n-k\text{ odd}\\\frac{2\pi}{2^k}\binom{k}{\frac{n+k}2}&\text{otherwise}
\end{cases}$$
We will assume $n$ is non-negative, since $a_n=a_{-n}.$
Thus, letting $i=\frac{k-n}{2},$ this becomes:
$$a_n=\sum_{i=0}^{\infty}\frac{1}{(n+2i)!}\binom{n+2i}{n+i}=\sum_{i=0}^{\infty}\frac1{i!(n+i)!}$$
A: $$
\begin{align}
e^ze^{\frac1z}
&=\sum_{k=0}^\infty\frac{z^k}{k!}\sum_{j=0}^\infty\frac1{j!z^j}\\
&=\sum_{k=-\infty}^\infty z^k\color{#0000F0}{\sum_{j=0}^\infty\frac1{(k+j)!j!}}\\
&=\sum_{k=-\infty}^\infty\color{#0000F0}{I_{|k|}(2)}\,z^k
\end{align}
$$
where it is convention that $\frac1{k!}=0$ when $k<0$.
$I_n$ is the Modified Bessel Function of the Second Kind.

Why is the Sum in $\boldsymbol{j}$ Even in $\boldsymbol{k}$?
In the preceding, it is stated that
$$
e^{z+\frac1z}=\sum_{k=-\infty}^\infty I_{|k|}(2)z^k
$$
which implies that $\sum\limits_{j=0}^\infty\frac1{(k+j)!j!}$ is even in $k$; and in fact it is, because of the convention, mentioned above, that $\frac1{k!}=0$ when $k<0$.
For $k\ge0$, we have
$$
\begin{align}
\sum_{j=0}^\infty\frac1{(-k+j)!j!}
&=\sum_{j=k}^\infty\frac1{(-k+j)!j!}\tag{1a}\\
&=\sum_{j=0}^\infty\frac1{j!(j+k)!}\tag{1b}
\end{align}
$$
Explanation:
$\text{(1a)}$: the terms with $j\lt k$ are $0$ due to the convention
$\text{(1b)}$: substitute $j\mapsto j+k$
A: Assuming you are looking for the Laurent series around $z=0$: Since
$$
e^z = 1 + z + \frac{z^2}{2!} + \cdots = \sum_{k=0}^\infty \frac{z^k}{k!}
$$
for all $z \in \mathbb{C}$, we also have
$$
e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \cdots = \sum_{m=0}^\infty \frac{1}{m!z^m}
$$
for all $z\neq 0$. Multiplying the two series together we get the Laurent series we are looking for:
\begin{align}
e^{z+1/z} &= \sum_{n=-\infty}^\infty \left(\sum_{k-m=n} \frac{1}{k!m!} \right) z^n \\
&= \sum_{n=-\infty}^\infty \left(\sum_{m=0}^\infty \frac{1}{(m+n)!m!} \right) z^n
\end{align}
