Evaluating $\sum_{n=0}^{\infty} e^{itn^{2}}\frac{k^{n}}{n!} $

Question

I'm having some trouble finding the value that this infinite sum converges to $$\sum_{n=0}^{\infty} e^{itn^{2}}\frac{k^{n}}{n!} $$ Is there a standard trick to solve it? The $n^{2}$ term is making it difficult for me to put into the form of a geometric sum.