Showing that $\sum_{n=1}^{\infty}e^{-n^2 \pi y}$ $\leq $ $2e^{-\pi y}$ When $y \geq 1$. How can I show that $\sum_{n=1}^{\infty}e^{-n^2 \pi y}$ $\leq $ $2e^{-\pi y}$ When $y \geq 1$. On using the inequality with the integral I get: $\sum_{n=1}^{\infty}e^{-n^2 \pi y} \leq \int_{1}^{\infty}e^{-\pi y t}dt = \frac{e^{-\pi y}}{\pi y}$.  From here I can conclude that the summation have the desired upper bound?
 A: You can try the following approach. For reasons of personal preference I shall replace the $y$ in your notation with $\alpha \geqslant 1$ and introduce $a \colon=\mathrm{e}^{-\pi \alpha}$. It is clear that $0<a \leqslant \mathrm{e}^{-\pi}<\mathrm{e}^{-1}=\frac{1}{\mathrm{e}}<\frac{1}{2}$.
We can produce the following estimate:
$$\sum_{n=1}^{\infty}a^{n^2}=\sum_{n \in \mathbb{N}^{\times}}a^{n^2}=\sum_{m \in \left(\mathbb{N}^{\times}\right)^{\underline{2}}}a^m \leqslant \sum_{m \in \mathbb{N}^{\times}}a^m=\frac{a}{1-a},$$
which relies on the fact that for a sequence of positive reals, the sum of the associated series coincides with the sum of the sequence understood in the general sense of summation of arbitrary families in topological groups. The notation $\left(\mathbb{N}^{\times}\right)^{\underline{2}}=\left\{m^2\right\}_{m \in \mathbb{N}^{\times}}$ expresses the set of all nonzero square natural numbers and in order to obtain the sum of the right-most (geometric) series we relied on the condition that $a<1$.
This upper bound can be used to reach the desired conclusion as long as we manage to establish that in the given situation $\frac{a}{1-a} \leqslant 2a$, which amounts to claiming that $a \leqslant \frac{1}{2}$ and is indeed a condition we have already ascertained $a$ to satisfy.
This concludes the solution.
P.S. An equally valid manner of justifying the inequality $\displaystyle\sum_{n=1}^{\infty}a^{n^2} \leqslant \displaystyle\sum_{n=1}^{\infty}a^n$ is to simply the remark the term-wise comparison $a^{n^2} \leqslant a^n$ occurring between the two series in question, as $n^2 \geqslant n$ for any $n \in \mathbb{N}$ and $a<1$.
