According to my notes, the function
$$u(x,t)=\sum_{n=0}^{\infty} e^{(1-n^2)t}\cos{nx}$$ should be continuous on $(0,\pi) \times (0,\infty)$, but I'm unable to prove it.
We have
$$|u(x+h_1,t+h_2)-u(x,t)|\le \sum_{n=0}^{\infty} e^{(1-n^2)t}|e^{(1-n^2)h_2}\cos{n(x+h_1)}-\cos{nx}|,$$
but I can't find a way to bound this as $(h_1,h_2)\to (0,0)$. Maybe I'm missing some useful inequality ?