Limit of a bell-shaped summand

Relevant wiki: Riemann Sums

$\begin{aligned} L & = \lim_{n \to \infty} \frac 1{\sqrt n}\sum_{k \in \mathbb Z} \color{#3D99F6} e^{-\frac {k^2}n} & \small \color{#3D99F6} \text{Since }e^{-\frac {k^2}n} \text{ is an even function} \\ & = \lim_{n \to \infty} \frac 1{\sqrt n} \left(1 + 2\sum_{k=1}^n e^{-\frac {k^2}n} \right) & \small \color{#3D99F6} \text{Note that }\lim_{n \to \infty} \frac 1{\sqrt n} = 0 \\ & = \lim_{n \to \infty} \frac 2{\sqrt n} \sum_{k=1}^n e^{- \left(\frac k{\sqrt n}\right)^2} & \small \color{#3D99F6} \text{Using Riemann sums:} \\ & = 2 \int_0^\infty e^{-x^2} dx & \small \color{#3D99F6} \lim_{n \to \infty} \sum_{k=a}^b f\left(\frac kn \right) = \lim_{n \to \infty} \int_{\frac an}^{\frac bn} f(x) \ dx \\ & = \sqrt \pi \text{ erf }(\infty) & \small \color{#3D99F6} \text{Error function erf }(z) = \frac 2{\sqrt \pi }\int_0^z e^{-t^2} dt \\ & = \boxed{\sqrt \pi} & \small \color{#3D99F6} \text{Since erf }(\infty) = 1 \end{aligned}$