Help to understand an answer for the examination of normal convergence First, thanks for your understanding that we are not all good in mathematics but at least we are working hard to understand it.
I asked this question before but I didn't understand the received comments to help. 
My question was: For all $ x\in \mathbb{R} $, evaluate the normal convergence of the series $$\sum_{n \geq 2} f_n(x) = \sum_{n \geq 2} \frac{x e^{-nx}}{\ln(n)}$$ over $[m, \infty)$ s.t $m > 0$, to see if the function $f$ is continuous over $(0, \infty)$. 
My attempt to solve it
I thought about studying the behavior of the series of $n$ to see if converge and then to evaluate the convergence with $x$ so 
\begin{eqnarray}
\sum_{n \geq 2} \frac{x e^{-nx}}{\ln(n)} = x \sum_{n \geq 2} \frac{1}{e^{nx} \ln(n)} \leq x \sum_{n \geq 2} \frac{1}{e^{n} \ln(n)}
\end{eqnarray}
then I didn't know what to do more.
The comments to help suggested :
$x e^{-nx}$ reaches its maximum at $x=1/n$ (provided $1/n>m$) and at $x=m$ when $1/n \leq m$. So for sufficiently large $n$, the sup norm is $\frac{me^{−mn}}{\ln n}$ which is summable. Then $\sum_{n \geq 1/m} \frac{me^{−mn}}{\ln n}$is convergent, hence $\sum_{n \geq 2} \sup |f_n(x)|$ is also convergent.
What I don't understand is:  $1-$  I don't understand how the sup was found "although it is written but I don't understand it". $2-$ why it is summable and convergent,  and $3-$ how does this help us to conclude that the function f is continuous over $(0,\infty)$.
 A: The most common way to find the supremum of a smooth function is to find where the derivative is $0$.
if $f_n(x) = xe^{-nx}/\ln n$, then
$$f_n'(x) = \frac{(1-nx)e^{-nx}}{\ln n}$$
Setting that equal to $0$ and noting that $e^{-nx}/\ln n$ is never $0$, you get $1-nx = 0$ or $x = \frac 1n$. Because $e^{-nx}$ is always positive, the sign of $f_n'(x)$ is determined by the $(1-nx)$ factor. So it is positive if $x < \frac 1n$ and negative when $x > \frac 1n$. I.e., $f_n(x)$ is increasing to the left of $\frac 1n$ and decreasing to the right. So $x = \frac 1n$ is a maximum.
But your functions are restricted to the domain of $[m, \infty)$. If $\frac 1n$ is in that domain, then it is the maximum, and $$\sup |f_n(x)| = f_n\left(\frac 1n\right) = \frac{1}{ne\ln n}$$
If $\frac 1n$ is not in the domain, so $m > \frac 1n$, then remember that $f_n(x)$ is decreasing when $x > \frac 1n$, which is everywhere in $[m,\infty)$. So the maximum value in the domain in this case has to be at $x = m$ and is given by
$$\sup |f_n(x)| = f_n(m) = \frac{me^{-nm}}{\ln n}$$
Now, for all $n > \frac 1m$, we will always have $\frac 1n < m$. So regardless of the $m$ used, for the tail of the series, $\sup |f_n(x)| = f_n(m) = \frac{me^{-nm}}{\ln n}$, and it is only the tail that determines if the series converges or diverges. The front of the series (all terms with $n < N$ for some set $N$) are just a finite sum, which you can always do.
The series $\sum_n f_n(x)$ is normally convergent if $\sum_n \sup_x |f_n(x)|$ converges to a value $< \infty$. As mentioned, we can concern ourselves only with the tail, which is 
$$\sum_{n > 1/m,2} \frac{me^{-nm}}{\ln n} = m\sum_{n > 1/m,2} \frac{(e^{-m})^n}{\ln n}\le m\sum_{n > 1/m,2} a^n$$
where $a = e^{-m}$, since $\ln n > 1$ for $n \ge 3$. Now since $m > 0, a= e^{-m} < 1$. Therefore $\sum_n a^n$ converges. And because $\sum_{n > 1/m,2} \frac{me^{-nm}}{\ln n}$ is a positive series that is bounded above by $\sum_n a^n$, it must also converge.
Therefore $\sum f_n(x)$ is normally convergent.

As to your question "how does this help to show $f$ is continuous". If you open your textbook to the section about normally convergent functions and study it, you will find the answer to that question. Don't come here asking questions until you've looked hard at your textbook. It is your first resource for finding answers. At the very minimum, I expect that your textbook includes the result that normally convergent $\implies$ uniformly convergent.
