How to check if the series is analytic?

Question

I want to verify that the series:

$$ \sum_{m=0}^{\infty} \frac{(-z)^m}{m! \cos\left( \frac{\pi}{2}(m+s)\right)} $$ is convergent in $z \in \mathbb{C}\backslash(-1,0],$ and $-1<Re(s)<0$. I tried to apply The Weierstrass-M test but I can't figure out how can I verify this. Any help is appreciated.

Answer 1

For $s\in \Bbb{C-2Z+1}$ then $$f(z)=\sum_{m=0}^{\infty} \frac{(-z)^m}{m! \cos\left( \frac{\pi}{2}(m+s)\right)}$$ is entire

and $$F(s,z)=\sum_{m=0}^{\infty} \frac{(-z)^m}{m! \cos\left( \frac{\pi}{2}(m+s)\right)}$$ (converges locally uniformly thus) is analytic for $s\in \Bbb{C-2Z+1},z\in \Bbb{C}$.

Due doe $O((2m+1+s)^{-1})$ growth it has simple poles at the hypersurface $2m+1\times\Bbb{C}$.

Answer 2

Since $m$ varies over integers, $\cos\left(\frac{\pi}{2}(m+s)\right)$ is periodic. In particular, we can rewrite the summation as $$ \sum_{k=0}^{3}\sum_{m=0}^{\infty} \frac{(-z)^{4m+k}}{(4m+k)! \cos\left(\frac{\pi}{2}(4m+k+s)\right)} = \sum_{k=0}^{3} c_k \sum_{m=0}^{\infty} \frac{(-z)^{4m+k}}{(4m+k)!} $$ where $c_k = 1 / \cos\left(\frac{\pi}{2}(k+s)\right)$. Each of the series on the innermost summation on the RHS is clearly absolutely convergent, so we are done.