Let $n \in \mathbb{N}, f_n(x)=e^{-(x-n)^2}$ and $g(x)=\Bigg\{\begin{array}{ll} 1 & x=0 \\ \frac{1-e^{-x^2}}{x^2} & x \neq 0 \\ \end{array} $
You can assume that g is continuous and bounded with maximum in 0. Show that $$\sum_{n=0}^{\infty}g \circ f_n$$ converges uniformly as $n \rightarrow \infty $.
I have shown that $0\leq f_n(x) \leq 1 $ and that $f_n(x)\rightarrow 0$ as $n \rightarrow \infty$. Thus I figured, that I could investigate $$\frac{1-e^{-u^2}}{u^2} $$ However M test with $\frac{1-e^{-u}}{u^2} $ yields divergence. Hence, I must admit that I need help. Any advice, solutions or ideas would be greatly appreciated.
$g$ is decreasing and $0\le f_n(x)\le1$. This implies that $g\circ f_n(x)\ge g(1)=1-e^{-1}=0.632121\dots$ and that the series $\sum g\circ f_n$ does not converge.
Also, it is easy to see that $g\circ f_n(x)$ converges point wise, but not uniformly, to $1$.