Show that $\theta(z)=\sum_{m=-\infty}^{\infty}e^{-\pi m^2z}$ converges and is analytic in $z\in \mathbb C,Re(z)>0$
Possible proof: Let $Re(z)=x$
Firstly note that $|e^{-\pi m^2z}|=e^{-\pi m^2x}\leq e^{-|m|}$ for $|m|$ large enough i.e. we can always find $m$ s.t $\frac{1}{|m|}\leq \pi x$.
Since $\sum_{m=0}^{\infty} e^{-|m|}$ converges. This implies that $\sum_{m=0}^{\infty} e^{-\pi m^2z}$ which implies $\theta (z)$ converges.
To prove the analytical part it note that $e^{-\pi m^2z}$ is analytical for all $m\in \mathbb Z$. Therefore it suffices to show that it converges uniformly on compact sets i.e. suffices to show that $\forall \delta>0$ $\theta(z)$ converges uniformly on $z\in \mathbb C,Re(z)>\delta$.
Similarly as above we have that for $z\in \mathbb C,Re(z)>\delta$. $|e^{-\pi m^2z}|=e^{-\pi m^2x}\leq e^{-\pi m^2\delta}\leq e^{-|m|}$ for $|m|$ large enough.
This implies that $\sum_{m=0}^{\infty} e^{-\pi m^2z}$ converges uniformly on $z\in \mathbb C,Re(z)>\delta$ which implies $\theta(z)=\sum_{m=0}^{\infty} e^{-\pi m^2z}+\sum_{m=1}^{\infty} e^{-\pi m^2z}$ converges uniformly.
Is this proof right?