Is this equality correct
$$ \frac{ 0 + A \varepsilon e^{-a \varepsilon} + A 2 \varepsilon e^{-a2 \varepsilon} + \dots }{ A + A e^{-a\varepsilon} + A e^{-a2\varepsilon} + \dots } = \frac{\varepsilon}{e^{-a\varepsilon} - 1}, $$
where $A$ and $a$ are constants and $\varepsilon$ is some small number? It seems to me that there should be no minus in the exponent on the r.h.s.
EDITS: Edits were made to correct the third term in the denominator and to add a negative sign in the exponents.
REMARK: This question was answered very well by @dxiv, so I don't understand why it's closed. @dxiv's answer was very helpful, indeed, and that can be useful for other readers of this site as well.
Hint: simplify $A$ between the numerator and denominator, and let $e^{-a \varepsilon}=b\,$ then the LHS is:
$$ \frac{\varepsilon \,( b + 2 b^{2} + \dots)}{ 1 + b + b^{2} + \dots }$$
For $\,|b| \lt 1 \iff a \varepsilon \gt 0\,$ the series converge $\displaystyle\sum_{n \ge 0} n b^n = \frac{b}{(1-b)^2}$ and $\displaystyle\sum_{n \ge 0} b^n = \frac{1}{1-b}$.
[ EDIT ] To sum up the comments, the above gives:
$$ \frac{ 0 + A \varepsilon e^{-a \varepsilon} + A 2 \varepsilon e^{-a2 \varepsilon} + \dots }{ A + A e^{-a\varepsilon} + A e^{-a2\varepsilon} + \dots } = \frac{\varepsilon \,( b + 2 b^{2} + \dots)}{ 1 + b + b^{2} + \dots } = \frac{\varepsilon \,b}{1-b} = \frac{\varepsilon}{\cfrac{1}{b}-1} = \frac{\varepsilon}{e^{a \varepsilon}-1} $$
This differs from the posted RHS of $\;\displaystyle\frac{\varepsilon}{e^{\color{red}{\mathbf{-}}a \varepsilon}-1}\;$ but that must have been a typo in the source text, since with a negative exponent $\,e^{\color{red}{\mathbf{-}}a \varepsilon}-1 \lt 0\,$ while the LHS is always positive.