Convergence of $g(x)\cdot f(x)$

Question

Let $g(x)=\frac{1-e^{-x^2}}{x^2}$ for $x \neq 0$,$g(0)=1$ and $f(x)=e^{-(x-n)^2}$. You can assume that g(x) is continuous and bounded with maximum 1 in x=$0$. Show that $\sum_{n=1}^{\infty}g(x)\cdot f(x)$ converges uniformly on $\mathbb{R}$. I have been able to show that for $x \leq 0$: $$ \left|g(x)f(x)\right|\leq f(x) \leq e^{-n^2}$$ which converges. However, I am unable to use the M'test for x>0. Any ideas?

Answer 1

Look for maximum of $e^{2xn-x^2}$ then use that $g<1$ to get everything less then $h(n)e^{-n^2}$ for some $h$. I can tell you more if you want.