Exponential sum behaves like linear term for large $t$ I've done some calculations on interesting mathematical objects and came to the conclusion that they would behave nicely as expected if we would have that
$$t \sim  2\sum_{n=1}^\infty \exp(-\pi n^2/t^2)$$
for large $t$. Here $\sim$ means that difference is bounded by some power of $\log(t)$ times a constant. Is that actually true and how can we prove it?
EDIT: I just got an idea: We should roughly have
$$
2\sum_{n=1}^\infty \exp(-\pi n^2/t^2) \sim \int_{\mathbb R} \exp(-\pi x^2/t^2) dx= t.
$$
Now one still has to estimate the error we make.
 A: The Fourier transform of $f(x) = \exp\left(-\tfrac{\pi x^2}{t^2}\right)$ is $\widehat{f}(\omega) = t\exp\left(-\tfrac{t^2\omega^2}{4\pi}\right)$. So by using the Poisson summation formula, we have
$$\sum_{n = -\infty}^{\infty}f(n) = \sum_{k = -\infty}^{\infty}\widehat{f}(k)$$
$$\sum_{n = -\infty}^{\infty}\exp\left(-\dfrac{\pi n^2}{t^2}\right) = \sum_{k = -\infty}^{\infty}t\exp\left(-\dfrac{t^2k^2}{4\pi}\right)$$
$$1+2\sum_{n = 1}^{\infty}\exp\left(-\dfrac{\pi n^2}{t^2}\right) = t\left[1+2\sum_{k = 1}^{\infty}\exp\left(-\dfrac{t^2k^2}{4\pi}\right)\right]$$
$$2\sum_{n = 1}^{\infty}\exp\left(-\dfrac{\pi n^2}{t^2}\right) = (t-1) + 2t\sum_{k = 1}^{\infty}\exp\left(-\dfrac{t^2k^2}{4\pi}\right)$$
Since $k^2 \ge 2k-1$ for all integers $k \ge 1$, we have $$0 \le 2t\sum_{k = 1}^{\infty}\exp\left(-\dfrac{t^2k^2}{4\pi}\right) \le 2t\sum_{k = 1}^{\infty}\exp\left(-\dfrac{(2k-1)t^2}{4\pi}\right) = \dfrac{2t\exp(-\tfrac{t^2}{4\pi})}{1-\exp(-\tfrac{t^2}{2\pi})} \to 0$$ as $t \to \infty$, and thus, $$2\sum_{n = 1}^{\infty}\exp\left(-\dfrac{\pi n^2}{t^2}\right) \sim t-1$$
A: Thanks to the remark from Steven Clark I came up with a nice solution using the modular theta function
$$
\theta : \mathbb H \to \mathbb C,\quad
\theta(z) := \sum_{n \in \mathbb Z }e^{\pi i n^2 z}.
$$
With the transformation law
$$\theta\left(- \frac{1}{z}\right) = \sqrt{ \frac{z}{i} } \theta(z)$$
we obtain
$$
\sum_{n \in \mathbb Z} \exp \left( - \frac{\pi n^2}{t^2} \right)
= \theta(i/t^2)
 = t \theta(it^2) \sim t.
$$
Hence, we get
$$t-1 \sim  2\sum_{n=1}^\infty \exp(-\pi n^2/t^2).$$
