Evaluating infinte sum of gamma distributions

Question

I have a function which consists of an infinite series of a gamma distribution:

$f(z) = (\frac{1}{2}\sum_{k=1}^{\infty}\Gamma(k/2)(\sqrt{2}z)^k/k!)$

Using the definition of the gamma distribution I have rewritten it to:

$f(z) = (\frac{1}{2}\sum_{k=1}^{\infty}(k/2)!(\sqrt{2}z)^k/k!)$

However I cannot find an immediate way to reduce it to a computational expression. I have considered the solutions here and here. But I am unable to solve it still.

I am not used to working with infinite series, so any help is appreciated.

Answer 1

The correct identity is $\Gamma(k+1)= k! = k\Gamma(k)$, so $$f(z)=\frac 1 2 \sum_{k\geq 1} \frac{\Gamma\left(\frac k 2\right)}{\Gamma(k+1)} 2^{\frac k 2}z^k\tag{1}$$ $$\begin{split} f^\prime(z) &= \frac 1 2 \sum_{k\geq 1} \frac{k\Gamma\left(\frac k 2\right)}{\Gamma(k+1)} 2^{\frac k 2}z^{k-1}\\ &= \frac 1 2 \sum_{k\geq 1} \frac{\Gamma\left(\frac k 2\right)}{\Gamma(k)} 2^{\frac k 2}z^{k-1}\\ &= \frac 1 2 \frac{\Gamma\left(\frac 1 2\right)}{\Gamma(1)}\sqrt{2} + \frac 1 2\sum_{p\geq 1} \frac{\Gamma\left(\frac {p+1} 2\right)}{\Gamma(p+1)} 2^{\frac {p+1} 2}z^{p}\\ f^{\prime\prime}(z) &= \frac 1 2\sum_{p\geq 1} \frac{p\Gamma\left(\frac {p+1} 2\right)}{\Gamma(p+1)} 2^{\frac {p+1} 2}z^{p-1}\\ &= \frac{\Gamma(1)}{\Gamma(2)}+\frac 1 2\sum_{p\geq 2} \frac{\Gamma\left(\frac {p+1} 2\right)}{\Gamma(p)} 2^{\frac {p+1} 2}z^{p-1}\\ &= 1+\frac 1 2\sum_{q\geq 1} \frac{\Gamma\left(\frac {q} 2+1\right)}{\Gamma(q+1)} 2^{\frac {q} 2+1}z^{q}\\ &= 1+\frac 1 2\sum_{q\geq 1} \frac{\frac q 2\Gamma\left(\frac {q} 2\right)}{q\Gamma(q)} 2^{\frac {q} 2+1}z^{q}\\ &= 1 + \frac 1 2f(z) \end{split}$$ Solving this O.D.E. yields: $$f(z) = \alpha e^{\frac x{\sqrt{2}}} + \beta e^{-\frac x{\sqrt{2}}}-2$$ The initial conditions are $f(0) = 0$ and $f^\prime(0)=\frac{\Gamma\left(\frac 1 2\right)}{\Gamma(2)}\sqrt{2} = \sqrt{2\pi}$. Can you finish?