Show that for all $x\in\mathbb R$ the series$$F(x)=\sum_{n=0}^{\infty}f_n(x)=\sum_{n=0}^{\infty}\frac{n^2+x^4}{n^4+x^2}$$ converges to a continuous function.
My attempt: First restrict to $[-r,r]\subset\mathbb R$ and use Weierstrass' M-test (see page 12 of this note by T. Tao)
Take the sup-norm: $$||f_n||_{\infty}=\sup\left\{\frac{n^2+x^4}{n^4+x^2}:|x|\leq r\right\}=\frac{n^2+r^4}{n^4+r^2}$$ $F|_{[-r,r]}(x)$ converges uniformly to a continuous function if $\sum_{n=0}^{\infty}||f_n||_{\infty}$ converges, which is true by comparison: $$\frac{n^2+r^4}{n^4+r^2}\leq \frac{n^2+r^4}{n^4}\leq \frac{1}{n^2}+\frac{r^4}{n^4}$$
And the fact that $F(x)$ converges uniformly on any interval $[-r,r]$ means that $F(x)$ must be point-wise convergent on the whole real line. (Is this correct?)
Please help me verify or improve this proof, thanks!
The proof looks fine to me.
The last fact you mentioned is correct. Actually, for any $x\in\mathbb{R}$, we can choose $r>|x|$. Since $F$ converges uniformly on $[-r, r]$, and every term in the sum is continuous, $F$ is continuous on $[-r, r]$, thus is continuous at $x$.