I'm working through a textbook and one of the questions is as follows: For $n$, $\epsilon$, and $\rho$ positive, consider the equation $n + \epsilon = z + ne^{-z/\rho}$. Expand the exponential term to third order and deduce that in the limit $\epsilon \rightarrow 0^+$, $\zeta \approx 2\nu - \frac{2}{3} \nu^2/\rho$ where $0 < \nu = n-p = O(\rho^{1/2})$ and $\zeta$ denotes the unique root satisfying $0 < \zeta - \epsilon < n$.
My work for the problem is below: Expanding the aforementioned expression gets $$n + \epsilon \approx z + n\left(1 - \frac{z}{\rho} + \frac{z^2}{2\rho^2} - \frac{z^3}{6\rho^3}\right) \implies \epsilon \approx z + n \left( - \frac{z}{\rho} + \frac{z^2}{2\rho^2} - \frac{z^3}{6\rho^3}\right).$$ As $\epsilon \rightarrow 0$, this becomes $$0 \approx z + n\left( - \frac{z}{\rho} + \frac{z^2}{2\rho^2} - \frac{z^3}{6\rho^3}\right) \implies 0 \approx 1 + n\left( - \frac{1}{\rho} + \frac{z}{2\rho^2} - \frac{z^2}{6\rho^3}\right).$$ Simplification yields $$\frac{1}{n} - \frac{1}{\rho} + \frac{z}{2\rho^2} - \frac{z^2}{6\rho^3} = 0 \implies -6\rho^3 \left(\frac{1}{n} - \frac{1}{\rho} \right) - 3\rho z + z^2 = -6 \frac{\rho^2}{n}(\rho-n) - 3\rho z + z^2 \approx 0.$$ This has roots $$z =\frac{3\rho \pm \sqrt{9\rho^2 + 24\frac{\rho^2}{n}(\rho - n)}}{2} = \frac{3\rho \pm \rho \sqrt{9 + 24\frac{1}{n}(\rho - n)}}{2}.$$ This doesn't match up with the solution I'm supposed to reach, so I'm a bit confused where I'm going wrong.
