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? 
 A: 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 $[0,+\infty)$.
