Expand: $e^{\sin z}$ at $z = 0$ I need hints to find the Laurent expansion of $\displaystyle e^{\sin z}$ at $z = 0$ in a simpler way. I am getting double series which I can't simplify.
ADDED:: Can it be simpler that the below? I also need to find the radius of convergence.
$$\Large e^{\sin z} = e^{\sum_{k = 0}^\infty \frac{(-1)^nz^{2n+1}}{(2n+1)!}} = \sum_{n = 0}^\infty \frac{\small{B(1, -1/3!, \dots , (-1)^n/(2n+1)!)} z^n}{n!}$$
 A: Exponential formula. 
Supose
$$ f(x)=a_1 x+{a_2 \over 2}x^2+{a_3 \over 6}x^3+\cdots+{a_n \over n!}x^n+\cdots.$$
Then
$$\exp f(x)=e^{f(x)}=\sum_{n=0}^\infty {b_n \over n!}x^n,$$
where
$$b_n=\sum_{\pi=\left\{\,S_1,\,\dots,\,S_k\,\right\}} a_{\left|S_1\right|}\cdots a_{\left|S_k\right|},$$
and the index $\pi$ runs through the list of all partitions $\{S_1,\ldots,S_k\}$ of the set $\{1,\ldots,n\}$. (When $k=0$ the product is empty and so equals $1$.)
We have $a_1=-a_3=a_5=-a_7=\cdots=1$ and $a_2=a_4=a_6=\cdots=0$.
So suppose we want $b_4$.  There are $15$ partitions of $\{1,2,3,4\}$: one corresponding to $1+1+1+1$, six corresponding to $2+1+1$, three corresponding to $2+2$, four corresponding to $3+1$, and one corresponding to $4$.  In this case we can discard the ones containing even numbers, so we need only consider $1+1+1+1$ and $3+1$.  We get
$$
b_4 = a_1^4 + 4a_3 a_1 = 1 - 4 = -3.
$$
Similarly,
$$
\begin{align}
b_0 & & & =1 \\[6pt]
b_1 & = a_1 & & = 1 \\[6pt]
b_2 & = a_1^2 & & = 1 \\[6pt]
b_3 & = a_1^3 + a_3 & & = 1-1 = 0  \\[6pt]
b_4 & = a_1^4 + 4a_3 a_1 & & = 1 - 4 = -3 \\[6pt]
b_5 & = a_1^5 + 10a_3 a_1^2 + a_5 & & = 1 - 10 + 1 = -8 \\[6pt]
b_6 & = a_1^6 + 20 a_3 a_1^3 + 10 a_3^2 + 6a_5 a_1 & & = 6 - 20 + 10 + 1 = -3 \\[6pt]
b_7 & = a_1^7 + 35 a_3 a_1^4 + 70 a_3^2 a_1 + 21a_5 a_1^2 + a_7 & & = 1 - 35 + 70 + 21 - 1 = 56
\end{align}
$$
And so on.
The radius of convergence is $\infty$, because this is a composition of two entire functions, and is therefore an entire function.
A: Note that $\frac{\mathrm{d}}{\mathrm{d}x}e^{\sin(x)}=\cos(x)e^{\sin(x)}$. Equating the coefficients for the series on both sides, that is
$$
\sum_{k=0}^\infty(k+1)a_{k+1}x^k=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k} \sum_{k=0}^\infty a_kx^k
$$
we get the following recursion for the coefficients:
$$
a_{n+1}=\frac1{n+1}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{(2k)!}a_{n-2k}
$$
Since $a_0=1$, we get
$$
e^{\sin(x)}=1+x+\frac12x^2+0x^3-\frac18x^4-\frac1{15}x^5-\frac1{240}x^6+\frac1{90}x^7+\dots
$$
Not working from the sequence of coefficients, but just knowing that the radii of convergence of both $\sin(x)$ and $e^x$ are infinite, the radius of convergence for the composition will also be infinite.

Generalization on the Same Idea
To get a result similar to the Exponential formula cited by Michael Hardy, we can use the preceding idea to get a recursion for general $f$. Assuming that $f(0)=0$, and $\displaystyle f(x)=\sum_{k=1}^\infty f_kx^k$
we get the following recursion for $\displaystyle e^{f(x)}=\sum_{n=0}^\infty a_nx^n$:
$$
a_0=1\quad\text{and}\quad a_{n+1}=\frac1{n+1}\sum_{k=0}^n(k+1)f_{k+1}a_{n-k}
$$
