I am trying to find a holomorphic function $f(z)$ with Taylor series
$$ f(z) = \sum_{n=0}^{\infty}c_n z^n$$
where the Taylor series coefficients are given by
$$ c_n = \sum_{\text{even } i}^n \frac{1}{(i+1)!}$$
The partial sums should be related to the incomplete Gamma function by the relation
$$ \sum_{k=0}^n \frac{x^k}{k!} = \frac{\Gamma(n+1,x)e^x}{n!}$$
and setting $x=1$. The question is then roughly can the incomplete Gamma function $\Gamma(n,x)$ be realised as an nth derivative! I'm dreaming of ultimately something like an integral expression for $f(z)$.
Thanks
Best Answer
The function you are looking for is rather simple: \begin{align*} \sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 0}^{\left\lfloor {n/2} \right\rfloor } {\frac{1}{{(2k + 1)!}}} } \right)z^n } & = \sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 0}^n {\frac{1}{{(2k + 1)!}}} } \right)z^{2n} } + \sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 0}^n {\frac{1}{{(2k + 1)!}}} } \right)z^{2n + 1} } \\ & = (1 + z)\sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 0}^n {\frac{1}{{(2k + 1)!}}} } \right)z^{2n} } \\ & = (1 + z)\left( {\sum\limits_{n = 0}^\infty {\frac{1}{{(2n + 1)!}}z^{2n} } } \right)\left( {\sum\limits_{n = 0}^\infty {z^{2n} } } \right) \\ & = (1 + z)\frac{{\sinh (z)}}{{z(1 - z^2 )}} = \frac{{\sinh (z)}}{{z(1 - z)}}. \end{align*}