TITLE: Closed form of $\sum_{n=0}^{\infty} \frac{\cos(\alpha \, n)}{n!}$
QUESTION [6 upvotes]: How would I find the closed form of
$$ f(\alpha) = \sum_{n=0}^{\infty} \frac{\cos(\alpha \, n)}{n!}$$
My old pal Wolfram tells me that
$$ f(\alpha) = \mathrm{e}^{\cos(\alpha)}\cos(\sin(\alpha))$$
I've attempted writing out the first couple terms for different values of $\alpha$, but I haven't quite figured out how to arrive at the result.

REPLY [4 votes]: Consider
\begin{eqnarray*}
\sum_{n=0}^{\infty}  \frac{\cos (n\alpha)}{n!}+i \sum_{n=0}^{\infty} \frac{\sin (n\alpha)}{n!}.
\end{eqnarray*}
Now by De Moivre's theorem this equals
\begin{eqnarray*}
\sum_{n=0}^{\infty}  \frac{e^{(ni\alpha)}}{n!}= e^{e^{i\alpha}}=e^{\cos \alpha+i\sin \alpha} =e^{\cos \alpha} e^{+i\sin \alpha} =e^{\cos \alpha} (\cos (\sin \alpha)+i\sin (\sin\alpha))
\end{eqnarray*}
Now equate the real parts & we have
\begin{eqnarray*}
\sum_{n=0}^{\infty}  \frac{\cos (n\alpha)}{n!}=e^{\cos \alpha} \cos (\sin \alpha).
\end{eqnarray*}