I'm interested in upper bounding the following infinite series by exponential functions for x>1 (I would be fine with x>2 as well).

\begin{eqnarray*} \sum_{n=1}^{\infty}e^{-\frac{x^2n^2}{2}}\\ \sum_{n=1}^{\infty}x^2ne^{-\frac{x^2n^2}{2}}\\ \sum_{n=1}^{\infty}x^2(x^2n^2-1)e^{-\frac{x^2n^2}{2}}\\ \sum_{n=1}^{\infty}nx^2(x^2n^2-3)e^{-\frac{x^2n^2}{2}} \end{eqnarray*}

By plotting them it seems that something like the following holds for $x>\sqrt{6}$. Essentially first term multiplied by 1.05

\begin{eqnarray*} \sum_{n=1}^{\infty}e^{-\frac{x^2n^2}{2}}\le 1.05e^{-\frac{x^2}{2}}\\ \sum_{n=1}^{\infty}x^2ne^{-\frac{x^2n^2}{2}}\le 1.05x^2e^{-\frac{x^2}{2}}\\ \sum_{n=1}^{\infty}x^2(x^2n^2-1)e^{-\frac{x^2n^2}{2}}\le 1.05x^2(x^2-1)e^{-\frac{x^2}{2}}\\ \sum_{n=1}^{\infty}nx^2(x^2n^2-3)e^{-\frac{x^2n^2}{2}}\le 1.05x^2(x^2-3)e^{-\frac{x^2}{2}} \end{eqnarray*} I was wondering if there was a way of formalizing the above. For example is it possible to show that the right-hand side of the above inequalities minus the left-hand side has only one zero. Then verifying its location by plotting and deduce the inequalities above hold after that value.

Also in case it might be helpful the first function is essentially the theta function \begin{equation*} \vartheta_3(q)=\sum_{n=1}^\infty q^{n^2} \end{equation*}


The first inequality you can get from comparison to $\sum_{n=1}^{\infty}e^{-\frac{x^2n}{2}}$ and then looking at it as a geometric series. On the others you need to prove essentially (after some reduction) that $\sum_{n=1}^{\infty}n^2e^{-\frac{x^2n^2}{2}}$ can be bounded by the same thing or something similar. In each case your bound for $\sum_{n=1}^{\infty} a_nb_n^n$ is just $\frac{a_1b_1}{1-b_1}$ which looks a lot like the geometric series.

  • $\begingroup$ I suggest comparing $e^{-x^2n^2/2+2\ln(n)}$ and $e^{-x^2n/2}$. I just worked it out and it shows that you can use a strict comparison for $x$ sufficiently large. $\endgroup$ – Stephen Sturgeon Jul 29 '13 at 18:16

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