This is an additional exercise given in my Fourier analysis course.
Define $F(t)=\sum_{n=-\infty}^\infty (-1)^n e^{-2n^2t^2}, \,t>0$, Prove that
- $\lim_{t\to \infty}F(t)=1$.
- $\lim_{t\to 0}\bigg(F(t)-\frac{\sqrt{2\pi}}{t}e^{-\frac{\pi^2}{8t^2}}\bigg)=0$.
My Try
For Question 1, for $t>1$ the series is uniformly convergent, hence $\exists N, |\sum_{n=-\infty}^\infty (-1)^n e^{-2n^2t^2}-\sum_{n=-N}^N (-1)^n e^{-2n^2t^2}|<\epsilon, \forall t>1$.
Since $\sum_{n=-N}^N (-1)^n e^{-2n^2t^2}\to 1$ as $t\to \infty$, $\exists M, \forall t>M,|\sum_{n=-N}^N (-1)^n e^{-2n^2t^2}-1|<\epsilon$, hence $\forall t>M,|\sum_{n=-\infty}^\infty (-1)^n e^{-2n^2t^2}-1|<2\epsilon$.
But I am not sure how to solve Q2. Any suggestion is appreciated.