Let $$ u_n(x) = \frac{x^n}{1 + x + x^2 + \dots + x^{n^2}}, \\\\ f(x) = \sum\limits_{n=1}^{\infty}u_n(x). $$
I need to check whether $f(x)$ is continuous on $x \geq 0$
I came to: $$ u_n(x) = \begin{cases} 1 \big/ (n^2 + 1), & x = 1; \\\\ x^n (x-1) \big/ (x^{n^2 + 1} - 1), & x \in (-\infty, 1) \cup (1, +\infty). \end{cases} $$
I can show that $f(x)$ is continuous at $x = 1$:
$u_n(x)$ is continuous for all $n \geq 1$;
Weierstrass M-test: $\dfrac{1}{n^2 + 1} \leq \dfrac{1}{n^2}$ for all $n \geq 1$, $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^2}$ is uniformly convergent $\implies \sum\limits_{n=1}^{\infty} \dfrac{1}{n^2 + 1}$ is uniformly convergent;
There is a theorem: If $\sum g_n(x)=g(x)$ (uniformly on $D$) and each $g_n$ is continuous at a point $x_0$ of $D$, then $g(x)$ is also continuous at $x_0$. Hence, $f(x)$ is continuous at $x = 1$.
Also, I can show that $f(x)$ is pointwise convergent on $x \geq 0$. But I don't know what to do with $x \neq 1$.