An upper bound for $\sum_{n=1}^{\infty}e^{-n^2}$ I am trying to find a good upper bound for
$$\sum_{n=1}^{\infty}e^{-n^2}\approx 0.3863186024.$$
I know that 
$$\int_{x=0}^{\infty}e^{-x^2}\;dx=\frac{\sqrt{\pi}}2 \approx  0.8862269255.$$

Is it possible to prove an upper bound less than $1/2$?

 A: You can use the following estimate for $\sum_{n=1}^\infty e^{-n^2}$:
$$\sum_{n=1}^N e^{-n^2} < \sum_{n=1}^\infty e^{-n^2} <\sum_{n=1}^N e^{-n^2} + e^{-N^2}.$$
Indeed,
$$\sum_{n=1}^\infty e^{-n^2} = \sum_{n=1}^N e^{-n^2} + \sum_{n=N+1}^\infty e^{-n^2} \leq \sum_{n=1}^N e^{-n^2} + \sum_{n=N+1}^\infty e^{-N^2} e^{N-n} = \sum_{n=1}^N e^{-n^2} + \frac{e^{-N^2}}{e-1} <  \sum_{n=1}^N e^{-n^2} + e^{-N^2}.$$
This formula gives the value of the sum with absolute error $e^{-N^2}$. E.g., for $N=3$, we get that the value of the sum lies between $0.3863$ and $0.3865$; for $N=5$, we get that it lies between $0.38631860241$ and $0.38631860243$.
Update: To get the desired upper bound of $1/2$, plug in $N=2$. Then the sum is bounded by $e^{-1} + 2 e^{-4}<0.405$.
A: Consider that $\mathbb e^{-n^2}\le \mathbb e^{-n}$. And so, $$\sum_{n=1}^\infty \mathbb e^{-n^2}\le \sum_{n=1}^\infty \mathbb e^{-n}=\frac{1}{(\mathbb e-1)}$$
A: I'm a bit late to the party, but here's a way to get a better estimation :
Let $m\geqslant 1$
We will want for $r \geqslant0 :$
$(p+r+1)^m \geqslant (p+\chi+r) p^{m-1} \iff \chi \leqslant m(r+1)+\frac{C_2^m(r+1)^2}{p}+\frac{C_3^m(r+1)^3}{p^2}+...+\frac{(r+1)^m}{p^{m-1}}-r$
This inequality is obviously true if we pick the expression for the lowest value of r (i.e. $r=0$). 
So we set $\chi=m+\frac{C_2^m}{p}+\frac{C_3^m}{p^2}+...+\frac{1}{p^{m-1}}=\frac{(p+1)^m}{p^{m-1}}-p$
Hence,
$\sum_{k=p}^{+\infty} e^{-k^m} \leqslant e^{-p^m}+\sum_{r=0}^{+\infty} (e^{-p^{m-1}})^{p+\chi+r}=e^{-p^m}+\frac{(e^{-p^{m-1}})^{p+\chi}}{1-e^{-p^{m-1}}}=e^{-p^m}+\frac{(e^{-p^{m-1}})^{\frac{(p+1)^m}{p^{m-1}}}}{1-e^{-p^{m-1}}}=e^{-p^m}+\frac{e^{-(p+1)^m}}{1-e^{-p^{m-1}}}$
So, if $l=\sum_{k=1}^{+\infty} e^{-k^m}$
$l-\sum_{k=1}^{k=p-1} e^{-k^m}=\sum_{k=1}^{+\infty} e^{-k^m}-\sum_{k=1}^{k=p-1} e^{-k^m} =\sum_{k=p}^{+\infty} e^{-k^m} \leqslant e^{-p^m}+\frac{e^{-(p+1)^m}}{1-e^{-p^{m-1}}}$
$\iff l-\sum_{k=1}^{k=p} e^{-k^m}=\sum_{k=p+1}^{+\infty} e^{-k^m} \leqslant \frac{e^{-(p+1)^m}}{1-e^{-p^{m-1}}}$
Fixing m=2, we get :
$l-\sum_{k=1}^{k=p} e^{-k^2}\leqslant \frac{e^{-(p+1)^2}}{1-e^{-p}}$
So after $p \geqslant1$ iterations, the error is at most $\delta(p) =\frac{e^{-(p+1)^2}}{1-e^{-p}}$
Some values are :
$\delta(1) \approx 0.28974914093$
$\delta(2) \approx 1.42725616 \times 10^{-4}$
$\delta(3) \approx 1.18431534 \times 10^{-7}$
$\delta(4) \approx 1.4147056 \times 10^{-11}$
$\delta(5) \approx 2.3352577 \times 10^{-16}$
