I'm given to functions, $f_{n}(x)=e^{-(x-n)^2}$ and $g(x)= \begin{cases} \frac{1-e^{-x^2}}{x^2} & x \neq 0 \\ 1 & x=0 \end{cases}$. I have to prove that $$\sum_{n=0}^{\infty} g(x) \cdot f_n(x) $$ converges uniformly.
I can assume that $g(x)$ is continuous and limited with maximum in $x=0$. I've tried Weierstrass M-test, but I can't get it to work out.
I'll put it her as well:
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.
We want a max for $e^{2xn-x^2}$, the derivative is $(2n-2x)e^{2xn-x^2}$ which is zero when $x=n$, now since for $x<n$ is positive and then is negative it is a local max (is actually a global max since the function goes to zero at $\pm\infty$). then in what I wrote above you get that $|gf|\leq n e^{-n^2}$ which is convergent.