Why is $\psi(x) = \sum_{n=1}^{\infty} e^{-\pi n^2x} = O(e^{-\pi x})$

Question

We define $\psi(x) = \sum_{n=1}^{\infty} e^{-\pi n^2x}$.

Why is it that $\psi(x) = O(e^{-\pi x})$

EDIT: As $ x \to \infty$

(big-oh-notation)

I think we can assume that x is positive.

I get that each term in the sum is smaller than $e^{-\pi x}$, but there are infinitely many terms, so how is this justified?

Answer 1

You're right; adding up infinitely many $O(f)$ terms needn't yield an $O(f)$ function.

However in this case the result does hold, and can be proven in a low-tech way. Hint:

$$\sum_{n=1}^\infty e^{-\pi n^2x}\le e^{-\pi x}\sum_{n=0}^\infty e^{-\pi nx} $$

Answer 2

I suppose that you mean $\psi(x)=_\infty O(x)$ and to prove it we have $$\left|\frac{e^{-\pi n^2x^2}}{x}\right|\le e^{-\pi n^2},\quad \forall x\ge1 $$ so the series $$\sum_n \frac{e^{-\pi n^2x^2}}{x}$$ is uniformly convergent on the interval $[1,\infty)$ since the series $\sum_n e^{-\pi n^2}$ is convergent hence $$\lim_{x\to\infty} \sum_n \frac{e^{-\pi n^2x^2}}{x}=\sum_n \lim_{x\to\infty}\frac{e^{-\pi n^2x^2}}{x}=0$$ hence the result follows and more precisely we have $$\psi(x)=_\infty o(x)$$