calculating $\lim_{n\to\infty}\frac{1}{n}\sum_{i=-\frac{nf}{2}}^{\frac{nf}{2}}e^{-(\frac{1}{n^2}+\alpha)i^2}$

Question

I am interested in calculating the following limit: \begin{equation} \lim_{n\to\infty}\frac{1}{n}\sum_{i=-\frac{nf}{2}}^{\frac{nf}{2}}e^{-(\frac{1}{n^2}+\alpha)i^2} \end{equation}

where $f$ and $\alpha$ are positive real-valued constants. does anybody have any idea how to proceed? any idea or approximation or even an asymptotic behavior would be welcomed.

i tried using Riemann sum formula for it but i failed since for the Riemann sum assuming $\Delta=\frac{b-a}{n}$the summand should have the form $f(\frac{m}{n})$ but as we see it doesn't have that form. for that reason i couldn't go any further using Riemann sum.

Answer 1

Let's just look at large enough $n$, such that $\frac 1n \ll\alpha$. Then the sum becomes (in the limit $n\to\infty$) $$S=\sum_{i=-\infty}^\infty e^{-\alpha i^2}$$ But this sum is just a little larger than the integral $$\int_{-\infty}^\infty e^{-\alpha x^2} dx=\sqrt{\frac{\pi}{\alpha}}$$ In fact the only thing that I care is that the sum is finite. When you divide by a large $n$, you get $0$.

Answer 2

We have

$$\frac{1}{n}\sum_{i=-\frac{nf}{2}}^{\frac{nf}{2}}e^{-\left(\frac{1}{n^2}+\alpha\right)i^2}=\frac{2}{n}\sum_{i=0}^{\frac{nf}{2}}e^{-\left(\frac{1}{n^2}+\alpha\right)i^2}-\frac{1}{n}$$

This can be compared to a geometric series

$$\sum_{i=0}^{\frac{nf}{2}}e^{-\left(\frac{1}{n^2}+\alpha\right)i^2}\leq \sum_{i=0}^{\frac{nf}{2}}e^{-\left(\frac{1}{n^2}+\alpha\right)i}=\frac{e^{\left(\frac{1}{n^2}+\alpha\right)}-e^{-\left(\frac{1}{n^2}+\alpha\right)\frac{nf}{2}}}{e^{\left(\frac{1}{n^2}+\alpha\right)}-1}<\frac{e^{\left(1+\alpha\right)}}{e^\alpha-1}$$

Then

$$0\leq \lim_{n\to\infty}\frac{1}{n}\sum_{i=-\frac{nf}{2}}^{\frac{nf}{2}}e^{-\left(\frac{1}{n^2}+\alpha\right)i^2}=\lim_{n\to\infty}\left[\frac{2}{n}\sum_{i=0}^{\frac{nf}{2}}e^{-\left(\frac{1}{n^2}+\alpha\right)i^2}-\frac{1}{n}\right]$$ $$\leq\lim_{n\to\infty}\left[\frac{2}{n}\frac{e^{\left(1+\alpha\right)}}{e^\alpha-1}-\frac{1}{n}\right]=0-0=0$$

Thus, the sequence converges to zero.