When can the sum of Gaussian-shaped terms over integers be approximated by the corresponding Gaussian integral

Question

I have the following finite sum: $$\frac{1}{\sqrt{n/12}}\sum_{k\in I} \exp\left(-\frac{(\frac{k}{\sqrt{n/12}} + \frac{1}{2\sqrt{3}})^2}{2}\right),$$ where$I$ is the interval $\{k\in \mathbb Z: -10n^{3/4} \leq k \leq O(1) -\frac{\sqrt{n}}{24}\}$.

My question is, can I write this sum approximately as an integral such that the identity $$\frac{1}{\sqrt{n/12}}\sum_{k\in I} \exp\left(-\frac{(\frac{k}{\sqrt{n/12}} + \frac{1}{2\sqrt{3}})^2}{2}\right)\\=\int_{-\infty}^{-1/(2\sqrt{12})} \exp\left(-\frac{(\frac{x}{\sqrt{n/12}} + \frac{1}{2\sqrt{3}})^2}{2}\right)dx\left(1 + O(g(n))\right)$$ holds? Here $g(n)$ is any suitable function that catches the asymptotic behavior nicely.

As an additional information. The guess of a professor of mine is, that this is "just an Riemann sum". But I dont see it clearly and I dont know how to get the asymptotics that are needed for the equation.

Edit: I have heavily edited the question. This makes the question more complex but more rigorous, too.

I am able to check the equality numerically in python. So it seems to be true. Here, and generally, too, we have $O(g(n)) = O(1/n)$.

I am happy for any suggestion! You do not have to give a proof. Any tips how it could be done are helpful!

Answer 1

I think there is still some mismatch in the $\frac{1}{\sqrt{n/12}}$ terms. As I understand it this term multiplies $k$ in the sum but not $x$ in the integral. My answer proceeds on this basis.

Let $$f(x) = \exp\left(-\frac{\left(x + \frac{1}{2\sqrt{3}}\right)^2}{2}\right)$$

Let $\ell = \min I, L=\max I$ and $a=-1/(2\sqrt {12})$.

So basically you are comparing

$$\sum_\ell^{L} \frac{1}{\sqrt{n/12}} f\left(\frac k {\sqrt{n/12}}\right) \mathrm{~with~} \int_{\infty}^a f(x) dx$$

Let us consider one small piece of this comparison:

$$\left|\frac{1}{\sqrt{n/12}} f\left(\frac k {\sqrt{n/12}}\right) - \int_{\frac{k}{\sqrt{n/12}}}^{\frac{k+1}{\sqrt{n/12}}} f(x) dx\right| \\ =\int_{\frac{k}{\sqrt{n/12}}}^{\frac{k+1}{\sqrt{n/12}}} \left| f\left(\frac k {\sqrt{n/12}}\right)-f(x)\right| dx $$

Now it is easy to see that $f'(x) < 1 \quad \forall x\in \mathbb R$. In fact, the derivative of $\exp(-x^2/2)$ attains it maximum value of $\sqrt{\exp(-1)}$ at $x=1$. Therefore the integrand is less than $\frac{1}{\sqrt{n/12}}$ and since the interval of integration is also $\frac{1}{\sqrt{n/12}}$, the absolute value of the integral is less than $\frac{12}{n}$

Since the number of elements in $I$ is $O(L-\ell) = O(n^{3/4})$, we get $$\left|\sum_\ell^{L} \frac{1}{\sqrt{n/12}} f\left(\frac k {\sqrt{n/12}}\right) - \int_{\frac{\ell}{\sqrt{n/12}}}^{\frac{L+1}{\sqrt{n/12}}} f(x) dx\right| = O(n^{3/4})\frac{12}{n} = O(n^{-1/4}) $$

It remains to consider the two tails of the integral. The left tail is very easy. The normal distribution tail decays exponentially:

$1-\mathfrak N(x) < x^{-1}\mathfrak n(x)= O(\exp(-x^2/2))$

Where $\mathfrak N(x)$ denotes the normal cumulative distribution function and $\mathfrak n(x)$ denotes the normal density. It follows that $\int_{-\infty}^{-\ell/\sqrt{n/12}} f(x) dx$ is negligible relative to the $O(n^{-1/4})$ term that we already have.

The final term is the right tail

$$\int_{a}^{\frac{L+1}{\sqrt{n/12}}} f(x) dx$$

Observe that

$$\frac{L+1}{\sqrt{n/12}}= \frac{-\sqrt n/24+O(1)}{\sqrt{n/12}} = a + O(n^{-1/2})$$

Since $|f(x)| \le 1 \quad \forall x \in \mathbb R$, the right tail integral is $O(n^{-1/2})$

Adding all three error terms, we get the total difference between the sum and the integral is $O(n^{-1/4})$.

If we let S be the sum and I the integral, we have proved that $|S - I| = O(n^{-1/4})$. From this, we can easily get the expression that you want with $g(n)=n^{-1/4}$. The integral $I$ is independent of $n$ and is therefore $O(1)$. In fact, the paper that you linked shows that it is $\Phi(a)$ where $\Phi$ denotes the normal cumulative distribution function. So $$S= I + (S-I)\\ = I + O(n^{-1/4})\\ = I + I\cdot O(n^{-1/4})/I \\ = I + I\cdot O(n^{-1/4})/O(1) \\ = I + I\cdot O(n^{-1/4}) \\ = I \cdot (1 + O(n^{-1/4})) $$