3
$\begingroup$

While reviewing the integral test, I remembered a formula for bounding sums the test applies to (non-negative, monotone decreasing $f(x)$): $$\sum_{N+1}^M f(n) \leq \int_N^M f(x)dx$$

Specifically, for $S_N = \sum_{k=1}^N e^{-k^2}$, $$S_\infty = \sum_{k=1}^\infty e^{-k^2} \leq \int_0^\infty e^{-x^2}dx = \frac{\sqrt{\pi}}{2}\operatorname{erf}(\infty) = \frac{\sqrt{\pi}}{2}$$

I computed $\frac{\sqrt{\pi}}{2}\operatorname{erf}(N) - S_N$ for large $N$ ($10^6$), and it seemed to converge to ~$0.49990832303943183$. This implies that $$\sum_{k=1}^\infty e^{-k^2} = \frac{\sqrt{\pi}}{2} - C,$$ where $C$ is ~$0.49990832303943183$. ries gives a few ideas for $C$, and also for $S_{10^6}$.

Can these estimates get any better? Or, fingers crossed, is there a closed form for this sum? Specifically without Jacobi Theta functions.

$\endgroup$
1
$\begingroup$

The only “closed” form of this sum is in terms of Jacobi theta functions. A better approximation (involving the erfc function) can be found using partial summation $$\sum_{k\leq N}e^{-k^{2}}=Ne^{-N^{2}}+2\int_{1}^{N}\left\lfloor t\right\rfloor te^{-t^{2}}dt $$ which means $$\sum_{k\geq1}e^{-k^{2}}=2\int_{1}^{\infty}t^{2}e^{-t^{2}}dt-2\int_{1}^{\infty}\left\{ t\right\} te^{-t^{2}}dt$$ where $\left\lfloor t\right\rfloor$ and $\left\{ t\right\} $ are the integer part of $t$ and the fractional part of $t$. Hence we have, using integration by pats, $$ 2\int_{1}^{\infty}t^{2}e^{-t^{2}}dt=\frac{1}{e}+\frac{\sqrt{\pi}}{2}\textrm{erfc}\left(1\right)\approx0.50782$$ so, noting that $\left\{ t\right\} \in\left[0,1\right)$, we have $$0\leq2\int_{1}^{\infty}\left\{ t\right\} te^{-t^{2}}dt\leq\frac{1}{e}$$ $$\sum_{k\geq1}e^{-k^{2}}=\frac{1}{e}+\frac{\sqrt{\pi}}{2}\textrm{erfc}\left(1\right)-C $$ with $C\in\left[0,1/e\right]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.