How to prove lower and upper bound for exponential sum?

Question

A paper I'm reading implicitly uses the fact $$\sum\limits_{t=1}^n e^{-ta^2} \in \theta(\frac{1}{a^2})$$ (It uses the both $\leq$ and $\geq$ sides in the proofs). I'm able to prove that $\sum\limits_{t=1}^n e^{-ta^2} \leq \frac{1}{a^2}$ by bounding the summation using integral. How can I prove $$\sum\limits_{t=1}^n e^{-ta^2} \geq c\frac{1}{a^2}$$ for some constant $c$?

Answer 1

By the geometric series: $$ \sum_{t=1}^n e^{-t a^2} = e^{-a^2}\frac{1-e^{-na^2}}{1-e^{-a^2}} $$ If $a$ is very large, this is approximately $e^{-a^2}$, and there is no way this is bounded below by $c/a^2$. And if $a$ is very small, this is approximately $n-1$ - again no way is this bounded below by $c/a^2$.