I am wondering if the following series has a closed form expression:
$\sum_{x=0}^{\infty} \frac{g(x)\lambda^x}{x!}$, where $g(x)=\frac{x}{2}$ for even $x$, and $g(x)=\frac{x-1}{2}$ for odd $x$.
I have tried splitting the series into one odd and one even series, but this doesn't seem to simplify anything.
For $g(x) = \frac{x}{2}$ we have
$$\frac{1}{2}\sum_{x=0}^\infty \frac{x\lambda^x}{x!} = \frac{\lambda}{2}\sum_{x=1}^\infty \frac{\lambda^{x-1}}{(x-1)!} = \frac{\lambda}{2}e^\lambda$$
which makes the other sum for $g(x) = \frac{x-1}{2}$
$$ \frac{\lambda-1}{2}e^\lambda$$
by exploiting the linearity of summations.
