Consider the function,
$$\delta(n)=\lim_{z\rightarrow\infty}\left(\sum_{n=1}^{\infty} e^{-z\left(x-n\right)^2}\right).$$
I am trying to prove that \begin{equation} \delta(n)=\begin{cases}1 & \text{if }n\in\mathbb{N}+1\\ 0 & \text{if $n\notin\mathbb{N}+1$}\end{cases}. \end{equation}
Theory of distributions might help, since the function is a quasi-Dirac delta function (but with impulse of magnitude one instead of infinity). The following property seems to be true, although I'm not sure of its rigour: \begin{equation} \int_{-\infty}^{\infty} \phi(x)\delta\left(x-a\right)\equiv0. \end{equation}
Any help would be much appreciated.
The Theory of distributions is of little help here, as the function we're studying is not continuous, and zero almost everywhere. The easiest way to prove your result is still to prove it pointwise. Let $x\in\mathbb R$.
If $x\in\mathbb N +1$, then for $z>0$, \begin{split} \sum_{n=1}^\infty e^{-z(x-n)^2} &= 1+\sum_{n=1}^{x-1} e^{-z(x-n)^2}+\sum_{n=x+1}^\infty e^{-z(x-n)^2} \\ &= 1+\sum_{n=1}^{x-1} e^{-z(x-n)^2}+\sum_{n=1}^\infty e^{-zn^2} \end{split}
The first sum in the right-hand side converges to zero as $z\to\infty$, and so does the second one by uniform convergence. Therefore $\delta(x)=1$.
If $x\neq\mathbb N +1$, then each of the exponents is non-zero so uniform convergence works as well to get $\delta(x)=0$.