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.
Even better, just as a quick estimate, one has right off the bat $$e^{\pi x}-1=\pi x + (\pi x)^2/2 + O(x^3)\gt(e^{\pi}-1)x, \forall x\gt1.$$ This inequality holds because the Taylor expansion of $e^{\pi x}$ has an infinite radius of convergence with each term of the series a monotonically increasing positive function of $x$. Therefore, since it has already been demonstrated that $$\sum_{n\ge 1}e^{- \pi n^2 x} < \frac{1}{(e^{\pi x}-1)}, \forall x>1$$ we have $$\sum_{n\ge 1}e^{- \pi n^2 x} < \frac{1}{(e^{\pi}-1)x} < \frac{1}{22 x} < \frac{1}{2 \sqrt x}, \forall x>1.$$
