Asymptotics of $\sum_{n}e^{-n^{2}}$.

Question

Define the function $S(N)$ as $$S(N)=\sum_{n=0}^{N}e^{-n^{2}}$$ I am interested in the asymptotic behavior of $S(N)$ for large $N$. It is clear by the ratio test that $\lim_{N\rightarrow\infty}S(N)$ exists.

One approach is to treat $S(N)$ as a Riemann sum for the function $e^{-x^{2}}$. Thus, define $I(N)$ as $$I(N)=\int_{0}^{N}e^{-x^{2}}\,dx$$ $I(N)$ is the error function $\textrm{erf}(N)$, for which the asymptotics are well known.

Is it valid to approximate $S(N)\approx I(N)$ for large $N$? This would immediately yield the asymptotic behaviour for $S(N)$.

Is there a direct way to investigate the asymptotics of $S(N)$?

Answer 1

Say $s=\sum_{n=1}^\infty e^{-n^2}$. It seems clear that $$s-S(N)\sim e^{-(N+1)^2}\quad(N\to\infty).$$

Right: $$\frac{s-S(N)}{e^{-(N+1)^2}}=\sum_{k=0}^\infty e^{-(2k(N+1)+k^2)} \to1\quad(N\to\infty),$$by dominated convergence, or by noting that last sum is dominated by a certain geometric series (throw away the $k^2$) or whatever. Heh, in fact $$0\le\frac{s-S(N)}{e^{-(N+1)^2}}-1 \le\sum_{k=1}^\infty e^{-2k(N+1)}=\frac{e^{-2(N+1)}}{1-e^{-2(N+1)}}\sim e^{-2(N+1)}.$$

Answer 2

By calculating just a handful of terms, one finds that

$$\sum_{n=0}^\infty e^{-n^2}\approx1.3863186$$

whereas

$$\int_0^\infty e^{-x^2}dx={\sqrt\pi\over2}\approx0.886227$$

so the "approximation" $S(N)\approx I(N)$, I'd say, is not valid for large $N$.