Can k (proportionate growth rate) be interpreted as $e^k-1$ (the proportion added/removed each period)? I am confused between:

*

*k: the proportionate growth rate

and

*

*$e^k-1$: the proportion added/removed each unit period

For example, if I have a population following the exponential growth function:
$P = P_0e^{0.04t}$ (so $k= 0.04$ year$^{-1}$)
and I am asked the proportion of the population which is added each year it's:
$\frac{P-P_0}{P_0}= e^k-1$
In numbers:
$e^{0.04}-1= 0.040810...$
So it's very close to $k$ but not quite. I would interpret this as "the population is growing by 4.08% each year".
This confused me because it's exactly what I thought the factor $k$ was meaning: the growth rate of a population. And indeed, in number they are very close ($k= 0.04$ and $e^k-1= 0.040810...$)
 A: You can write $e^{0.04^t}$ as $\lim\limits_{n\to \infty}\left(1+\frac{0.04}{n}\right)^{n\cdot t}$. This is a continuous growth.
For $t=1$ (one year) we have $e^{0.04}=\lim\limits_{n\to \infty}\left(1+\frac{0.04}{n}\right)^{n}$. Here we have $n$ growths per year with an growth rate of $\frac{0.04}{n}$, where $n$ goes to infinity.
Now we can look what the growth rate is when the number of growth intervals do not go to infinity, but it can be  large: $\left(1+\frac{0.04}{n}\right)^{n}$
We can use the binomial theorem to obtain $\left(1+\frac{0.04}{n}\right)^{n}=\color{blue}{1+0.04}+\binom{n}{2}\cdot \left(\frac{0.04}2\right)^2+...$
We see for every $n>1$ it is greater than $1.04$ and only for $n=1$ we have an equality :$\left(1+\frac{0.04}{1}\right)^{1}=1.04$.
Conclusion: We can divide our growth period into $n$ growth sub-intervals with an growth rate $\frac{r}n$. But then the growth rate of the overall period is greater than $r$. In the case of $n\to \infty$ the growth rate of the overall period becomes $e^r-1>r$.
