# $\sum \exp(-x^2)$ vs $\sum x^2 \exp(-x^2)$

I am curious about the following sum, for $$\alpha \in (0,1)$$:

$$\sum_{k = -\infty}^{\infty} (1-(2k - 1 + \alpha)^2) \exp(-\frac{1}{2} (2k - 1 + \alpha)^2)$$

I have reasons to believe sum should be zero when $$\alpha = 1/2$$, but I don't know how to prove it. And according to Mathematica, the value at $$\alpha = 1/2$$ is

$$-2.6474039 \times 10^{-7}$$,

which is notably not zero (though I'm not convinced that this isn't a rounding error). So: prove me wrong! (Is it possible to show that the value of this sum is non-zero?)

For some 'intuition' about why the sum is so close to zero, note that

$$\int_{-\infty}^{\infty} (1-(2x - \alpha + 1)^2)\exp(-\frac{1}{2} (2x - \alpha + 1)^2) \, dx = 0$$

for all $$\alpha \in \mathbb{R}$$. This is because the value of the integral doesn't depend on $$\alpha$$, and computing the value at $$\alpha = 0$$ is an easy exercise. The sum above is a Riemann sum for this integral. Of course, there is always error in changing from a Riemann sum to an integral -- I am just suspicious of how small the error is!

I would also like to know if there is a 'nicer' expression for the sum, or some relationship to well-known special functions. I suspect there are some clever Fourier-analysis type ideas to help evaluate this kind of sum.

• These are related to the Jacobi theta function. I'd try Poisson summation. – Lord Shark the Unknown Nov 25 '18 at 7:12

## 1 Answer

Let $$f(k) = \sum_{k = -\infty}^{\infty} (1-(2k - 1 + \frac12)^2) \exp(-\frac{1}{2} (2k - 1 + \frac12)^2)$$, so that you are asking about the sum $$S = \sum_{k=-\infty}^\infty f(k)$$.

We can calculate that $$\sum_{k=-2}^3 f(k) \approx -2.37104\times10^{-7}$$ (and presumably you are comfortable trusting this finite calculation). This proves that $$S<0$$, since the omitted summands $$f(k)$$ are negative for all $$k\le-3$$ and all $$k\ge4$$.

Indeed, we can also calculate $$\sum_{k=-4}^5 f(k) \approx -2.647403940047578\times10^{-7} \tag{*}$$ and, since $$|f(t)|$$ is increasing for $$t<-2$$ and decreasing for $$t>2$$, $$\sum_{k=6}^\infty |f(k)| < \int_5^\infty |f(t)|\,dt = |F(5)| < 1.2\times10^{-19}$$ and $$\sum_{k=-\infty}^{-5} |f(k)| < \int_{-\infty}^{-4} |f(t)|\,dt = |F(-4)| < 8.7\times10^{-16},$$ where $$F(t) = \int f(t)\,dt = \frac{1}{4} e^{-2t^2+t-\frac{1}{8}} (4t-1);$$ these calculations show that the right-hand side of $$(*)$$ is the correct value of the infinite sum $$S$$ to 7 or 8 significant digits.