Normal convergence: $\sum_{n = 1}^{\infty} \frac{x}{(1 + (n-1)x)(1 + nx)}$ I want to prove that:
$$\sum_{n = 1}^{\infty} \frac{x}{(1 + (n-1)x)(1 + nx)}$$
is not normally convergent on $[a, \infty)$ for fixed $a>0$.
Let $U_n(x)$ denote the general term.
We have:
$$||U_n||_{\infty} = \sup_{x \ge a} \left( \frac{x}{(1 + (n-1)x)(1 + nx)} \right) \ge \sup_{x \ge a} \left( \frac{x}{(1 + (n-1)a)(1 + na)} \right) \ge \sup_{x \ge a} \left( \frac{x}{(1 + na)^2} \right) \ge \frac{1}{(1 + a)^2} \sup_{x \ge a} \left( \frac{x}{n^2} \right) \ge \frac{1}{(1+ a)^2} \times \frac{n}{n^2} = \frac{1}{(1+a)^2} \frac{1}{n} = a_n$$
The series of general term $a_n$ is divergent, so by comparison test the series of norms is divergent.
Can someone check/criticize my work? Thanks.
Bonus question: what are the sets on which we have normal convergence? 

Your first inequality is invalid, if you make the (positive) denominator smaller, since the numerator is also positive, you make the fraction larger.
Actually, the series is normally convergent on $[a,+\infty)$ for all $a > 0$, since for $n \geqslant 2$
$$0 < U_n(x) = \frac{x}{(1+(n-1)x)(1+nx)} < \frac{x}{(n-1)xnx} = \frac{1}{n(n-1)x} \leqslant \frac{1}{n(n-1)a},$$
and
$$0 < U_1(x) = \frac{x}{1+x} < 1.$$
The series $\sum_{n=2}^\infty \frac{1}{n(n-1)}$ is convergent, so
$$\sum_{n=1}^\infty \lVert U_n\rVert_\infty \leqslant 1 + \frac{1}{a}\sum_{n=2}^\infty \frac{1}{n(n-1)} = 1+\frac{1}{a} < +\infty.$$
If there is a problem, it occurs at $0$. For $n \geqslant 2$, we compute
$$U_n'(x) = \frac{(1+(n-1)x)(1+nx) - x[(n-1)(1+nx) + n(1+(n-1)x)]}{[(1+(n-1)x)(1+nx)]^2}$$
and to locate the maximum solve
\begin{align}
&&(1+(n-1)x)(1+nx) &= x[2n-1 + 2n(n-1)x]\\
&\iff& 1 + (2n-1)x + n(n-1)x^2 &= (2n-1)x + 2n(n-1)x^2\\
&\iff& 1 &= n(n-1)x^2\\
&\iff& \frac{1}{\sqrt{n(n-1)}} &= x,
\end{align}
where the last is an equivalence since we only consider $x \geqslant 0$. Thus
\begin{align}
\lVert U_n\rVert_\infty &= U_n \biggl(\frac{1}{\sqrt{n(n-1)}}\biggr) \\
&= \frac{1}{\Bigl(1 + \sqrt{\frac{n-1}{n}}\Bigr)\Bigl(1 + \sqrt{\frac{n}{n-1}}\Bigr)\sqrt{n(n-1)}}\\
&= \frac{1}{(\sqrt{n} + \sqrt{n-1})(\sqrt{n-1}+\sqrt{n})}\\
&= \frac{1}{(\sqrt{n}+\sqrt{n-1})^2}\\
&= \frac{1}{2n-1 + 2\sqrt{n(n-1)}}.
\end{align}
Since $\lVert U_n\rVert_\infty > \frac{1}{4n}$, we see that the series is not normally convergent on all of $[0,+\infty)$.
