I am trying to show that
$$\sum_{n=1}^\infty e^{- \pi n^2 x} < \frac{1}{2} x^{-\frac{1}{2}}, \; \forall x>1$$
Here's what I'm doing:
$$\sum_{n=1}^\infty e^{- \pi n^2 x} < \sum_{n=1}^\infty e^{- \pi n x} = \sum_{n=0}^\infty e^{- \pi n x}-1=\frac{1}{1-e^{-\pi x}}-1 = \frac{1}{e^{\pi x}-1} <\frac{1}{2\sqrt{x}}, \; (x>1)$$
Is that correct?
Thanks in advance :)
Since $$\sum_{n=1}^{\infty}\left( \frac{1}{e^{\pi x}}\right)^{n}=\frac{1}{1-\frac{1}{e^{\pi x}}}$$
and
$e^{\pi x}-1=\pi x + (\pi x)^2 + O((\pi x)^3)>2\sqrt{x}$, $\forall x>1,$
the estimate is correct.