Let $m$ and $n$ be two positive integers greater than 1. If $\lim_{\alpha \to 0} \left( \frac{e^{\cos(\alpha^{n})} - e}{\alpha^{m}} \right) = -\left( \frac{e}{2} \right)$ then the value of $\frac{m}{n}$ is

$m, n \in \mathbb{Z}^{+},\qquad m > 1,\ n > 1$

$\lim_{\alpha\to 0} \dfrac{e^{\cos(\alpha^n)} - e}{\alpha^m} = -\dfrac{e}{2}$

$\text{To find: } \dfrac{m}{n}$

$\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$

$\cos(\alpha^n) = 1 - \dfrac{\alpha^{2n}}{2} + \dfrac{\alpha^{4n}}{24} - \cdots$

$e^{\cos(\alpha^n)} = e \cdot e^{\cos(\alpha^n)-1}$

$\cos(\alpha^n) - 1 = -\dfrac{\alpha^{2n}}{2} + \dfrac{\alpha^{4n}}{24} - \cdots$

$e^{\cos(\alpha^n)} = e\left(1 - \dfrac{\alpha^{2n}}{2} + o(\alpha^{2n})\right)$

$e^{\cos(\alpha^n)} - e = e\left(-\dfrac{\alpha^{2n}}{2} + o(\alpha^{2n})\right)$

$= -\dfrac{e}{2}\,\alpha^{2n} + o(\alpha^{2n})$

$\dfrac{e^{\cos(\alpha^n)} - e}{\alpha^m} = -\dfrac{e}{2}\,\alpha^{2n-m} + o(\alpha^{2n-m})$

$\text{Finite, nonzero limit} \ \Rightarrow\ 2n - m = 0 \ \Rightarrow\ m = 2n$

$m = 2n \ \Rightarrow\ \dfrac{m}{n} = 2$

$\boxed{\dfrac{m}{n} = 2}$