Find a general expression $c_n$ in the Laurent series
$$f(z)=\sum_{n=0}^\infty c_nz^n$$
for the function $f(z) = e^{z+1/z}$.
So I have
\begin{align} f(z) &= \sum_{i=0}^\infty \frac{z^i}{i!}\sum_{j=0}^\infty \frac{z^{-j}}{j!} \\ &= \sum_{i,j=0}^\infty \frac{z^{i-j}}{i!j!} \end{align}
Now we want the coefficient $c_n$ so set $i-j=n$ then we have that
$$f(z) = \sum_{i=0}^\infty\sum_{n=i}^\infty \frac{z^n}{i!(i-n)!}$$
Now I'm not really sure how to simplify this. Can anyone provide some hints?
Because the summation indices are non-negative, you have to split the summation on $n$ $$ \left\{ \matrix{ 0\, \le \,i \hfill \cr 0\, \le \,j \hfill \cr i - j = n \hfill \cr} \right.\quad \to \quad \left\{ \matrix{ 0\, \le \,j \hfill \cr 0\, \le \,n \hfill \cr i = n + j \hfill \cr} \right.\;\; \vee \;\left\{ \matrix{ 0\, \le \,i \hfill \cr n < 0 \hfill \cr j = i - n \hfill \cr} \right. $$ Therefore: $$ f(z) = \sum\limits_{0\, \le \,i,\,j} {{{z^{\,i - j} } \over {i!j!}}} = \sum\limits_{n\, < \,0} {\left( {\sum\limits_{0\, \le \,i} {{1 \over {i!\left( {i - n} \right)!}}} } \right)z^{\,n} } + \sum\limits_{0\, \le \,n} {\left( {\sum\limits_{0\, \le \,j} {{1 \over {j!\left( {n + j} \right)!}}} } \right)z^{\,n} } $$ to get the result provided in the first comment