Using the Weierstrass test, show that the series sum

Using the Weierstrass test, show that the series $\displaystyle\sum_{n=1}^{\infty}\frac{e^{-n^2x^2}}{n^2}$ converges uniformly on $(-\infty,\infty)$.

Step 1 · Uniform bound

For any $x\in\mathbb R$: $x^2\ge0\implies n^2x^2\ge0\implies e^{-n^2x^2}\le1$, so $$ \frac{e^{-n^2x^2}}{n^2} \le \frac1{n^2} $$ for every $x$ and $n\ge1$.

Step 2 · Majorant series

$\displaystyle\sum_{n=1}^{\infty}\frac1{n^2}=\frac{\pi^2}6<\infty$.

Step 3 · Weierstrass M-test

With $M_n=1/n^2$, both conditions hold, so the series converges uniformly and absolutely on $\mathbb R$. The sum $f(x)=\sum_{n=1}^{\infty}e^{-n^2x^2}/n^2$ is therefore continuous everywhere, with $f(0)=\pi^2/6$ (the maximum value, since each term is maximized at $x=0$) and $f(x)\to0$ rapidly as $|x|\to\infty$ due to the Gaussian factors.

Summary Table

Additional Notes

• This series is a close cousin of Jacobi theta functions $\theta(x)=\sum e^{-n^2x^2}$ (without the $1/n^2$ weight), which play a central role in analytic number theory and the functional equation of the Riemann zeta function; the extra $1/n^2$ factor here only makes convergence easier.
• Unlike Problems 194 and 197, where the majorant bound $1/n^2$ or $1/n^3$ came from discarding a non-negative additive term, here it comes from bounding a multiplicative Gaussian factor by $1$ — a different mechanism reaching the same M-test conclusion.
• Because $e^{-n^2x^2}$ decays extremely fast in $n$ for any fixed $x\ne0$, the true rate of uniform convergence is far better than the crude bound $1/n^2$ suggests away from $x=0$; the bound is tight only in a shrinking neighborhood of $x=0$, similar in spirit to the boundary-concentration phenomena seen in Problems 176 and 195.