The question is how can I show the following approximation stated in this paper.
I have tried to follow , but I am not pretty sure if the following results are correct
Generalized logistic growth modeling of the COVID-19 outbreak: comparing the dynamics in the 29 provinces in China and in the rest of the world
https://link.springer.com/article/10.1007/s11071-020-05862-6
Eq 5, 6
They have written an equation similar to
\begin{align} \frac{1}{Y}\frac{dY}{dt}=\frac{ A e^{- b t}}{ 1 - A e^{- b t}} \end{align} following the definition of the serie geometric
\begin{align} \sum_{n=0}^{\infty} Ae^{-n b t} = 1 + e^{- b t}+ e^{-2 b t} + e^{- 3b t} + \ldots \end{align}
\begin{align} \frac{1}{Y}\frac{dY}{dt}= \frac{ A e^{- b t}}{ 1 - A e^{- b t}} = A e^{- b t} \left( 1 + e^{- b t}+ e^{-2 b t} + e^{- 3b t} + \ldots\right) \end{align}
\begin{align} \frac{1}{Y}\frac{dY}{dt}= \frac{ A e^{- b t}}{ 1 - A e^{- b t}} = A e^{- b t} + e^{- 2b t}+ e^{-3 b t} + e^{- 4 b t} + \ldots \end{align} for long term, it can be written after substitution the approximation.
\begin{align} \frac{1}{Y}\frac{dY}{dt}= \frac{ A e^{- b t}}{ 1 - A e^{- b t}} = A e^{- b t} \end{align}
Any suggestion
$$\gamma \sum _{k=1}^{\infty } \left(\epsilon_0 e^{-\gamma t}\right)^k=\frac{\gamma \epsilon_0 e^{-\gamma t}}{1-\epsilon_0 e^{-\gamma t}}$$ In the sum on the left all ther terms with exponent $k$ larger that $1$ can be neglected for large $t$.
And it can be approximated by $\epsilon_0 e^{-\gamma t}$.
Just to see what happens, suppose $\epsilon_0 =\gamma=1$ and $t=5$, and let's see the values when exponent $k$ increases
$$ \begin{array}{r|r} k & e^{-5k}\\ \hline 1 & 0.006738 \\ 2 & 0.000045 \\ 3 & 3.06\cdot 10^{-7} \\ 4 & 2.06\cdot 10^{-9} \\ 5 & 1.39\cdot 10^{-11} \\ \end{array} $$