Question regarding the series $$\sum_{n=1}^{\infty} e^{-n^2x}=\sum_{n=1}^{\infty} f_n(x)$$. I wished to show that the sumfunction $f(x)=\sum_{n=1}^{\infty} f_n(x)$ is continuous on $]0,\infty[$ (it is readily discontiuous for $x<0$).
Now, I believe to have shown uniform convergence for any interval $x\in[a,\infty[$, with $a$ being some constant number $a>0$. To wit
$$e^{x}\geq x \forall x>0 \implies e^{n^2x} \geq n^2x \forall x>0 \iff \frac{1}{e^{n^2x}}\leq \frac{1}{n^2x}\forall x>0$$
So for $a>0$ I get $$\implies \frac{1}{e^{n^2x}}\leq \frac{1}{e^{n^2a}}\leq \frac{1}{n^2x} \leq \frac{1}{n^2a}$$ And $\sum_{n=1}^{\infty}\frac{1}{n^2a} = \frac{1}{a}\sum_{n=1}^{\infty}\frac{1}{n^2}$ is a convergent majorant series.
Now to the actual question... if a series of continuous functions is uniformly convergent, then its sumfunction is continuous (given in our textbook).
Can I simply say "pick any $a>0$, then $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly on $[a,\infty[$" since this holds for any arbitrary $a>0$, then the sumfunction is continuous on $]0,\infty[$.
This makes sense to me, but I'm not sure is stringent enough?
The answer is yes. If $g_n$ is a sequence of continuous functions such that, for every $a>0$, $g_n\to g$ uniformly in $[a,\infty)$, then $g$ is continuous in $(0,\infty)$.
To see that, let $y\in(0,\infty)$ and consider $a=y/2$. Since $g_n$ converges uniformly to $g$ in $[y/2,\infty)$ and $g_n$ is continuous in $[y/2,\infty)$, you automatically have that $g$ is continuous in $[y/2,\infty)$ (and in particular, $g$ is continuous at $y$).
Now take $g_n=\sum_{k=1}^n f_n$.