I was completely confused by that same explanation for a while. Don't even think of it like that. Think of it like this.
Definition: The effective rate of interest during the $n$th time period is
$$i_n = \frac{A(n) - A(n-1)}{A(n-1)}$$
Definition: The effective rate of discount during the $n$th time period is
$$d_n = \frac{A(n) - A(n-1)}{A(n)}$$
where $A(n)$ is the amount function (as defined in Kellison), the amount of money you have at time $n$. So, all you really need to understand here is that the rate of interest is a rate based on what you start with during the period. The rate of discount is a rate based on what you end up with. It's just two ways of looking at the same situation. There aren't a whole lot of real world situations where you borrow a bunch of money and immediately give some of it back. You would just borrow less.
By the way, that formula is all you need to calculate the effective rate of discount during period $n$ no matter what your $A(n)$ function is. So, in particular, it would work for your specific question of simple discount.
Question: Given a rate of 10% simple discount, calculate the effective rate of discount during period 5.
Answer: If we have 10% simple discount, then we know our accumulation function is $a(t) = \frac{1}{1 - 0.1t}$ for $0 \leq t < \frac{1}{d} = 10$. This is basically the definition of simple discount. If you have simple discount, this is your accumulation function. Memorize that. Then use it.
Therefore
$$d_5 = \frac{a(5) - a(4)}{a(5)} = \frac{2-10/6}{2} = \frac{1}{6} = 16.666666... \%$$
If you wanted to calculate the effective rate of interest when you are given the effective rate of simple discount, you can do that too. For example, in this same example,
$$i_5 = \frac{a(5) - a(4)}{a(4)} = \frac{2-10/6}{10/6} = \frac{1}{5} = 20 \%$$
Nothing changed. We're just looking at the same problem differently. In the discount case, how much money did we earn that period relative to how much we had at the end? In the interest case, how much money did we earn that period relative to how much we had at the beginning.
Note that in the United Kingdom where we have compound interest, consumer law requires APR and EAR interest rates to mean the same thing (apart perhaps from fees and other non-interest charges). I will use $r$ rather than your APR.
You are asking whether $(1+\frac rk)^k$ is an increasing function of positive $k$ (given fixed and presumably positive $r$)
and have found $f'(k)=(1+\frac rk)^k\left( \log_e(1+\frac rk) - \frac{r}{k+r}\right)$
which, as you note, is positive when $g(k)=\log_e(1+\frac rk) - \frac{r}{k+r} \ge 0$.
You could then say $\lim\limits_{k\to \infty} g(k)=0-0=0$ and $g'(k)=-\frac{r^2}{k(k+r)^2}<0$,
so $g(k)$ is decreasing towards $0$ as $k$ increases,
meaning $g(k)$ is positive for all finite positive $k$ and similarly $f'(k)$ is positive
and thus $f(k)$ is an increasing function of $k$.
Best Answer
Here you have a unit of time (month, year, for example). Call the unit of time the base period.
If these rates are to be equivalent (whether the investent is continuous, every unit of time or every $1/m$-th of a unit of time) then $$e^{rT}=(1+r_1)^T=(1+\tfrac{r_m}{m})^{mT}$$
The sequence $r_m$ also satisfies $r_m\xrightarrow{m\rightarrow\infty}r$. Indeed, $$r_m=m(e^{r/m}-1)=r\frac{e^{\tfrac{r}{m}}-1}{\tfrac{r}{m}}\xrightarrow{m\rightarrow\infty}r$$ since $\lim_{h\rightarrow0}\frac{e^h-1}{h}$ is the value of the derivative at $t=0$ of the function $f(t)=e^t$.
The convexity of the exponential function implies in fact that $r_m$ is monotone decreasing.