$$\mu_n = \frac{(b^{n-\gamma+1}-a^{n-\gamma+1})(-\gamma+1)}{(b^{-\gamma+1}-a^{-\gamma+1})(b-\gamma+1)}$$
I'm interested in how this behaves as $\gamma$ changes. We can assume $\gamma > 1$. I've played around with FOILing things out but have not really gotten anywhere. Hints appreciated!
If $n > 1$, then the limit is infinite. If $n < 1$, then the limit is zero. If $n = 1$, then the limit is $-\gamma + 1$.
Very generally, if you have two polynomials $p(x), q(x)$, then it is quite easy to compute the limit $$ \lim_{x \to \infty} \frac{p(x)}{q(x)}.$$ The term with larger degree wins. If the numerator has the same degree as the denominator, then the limit is the ratio of leading coefficients. This stays true even if the "degree" isn't an integer. We can easily describe the methodology of the proof.
Call $d$ the maximum of the degrees of $p(x), q(x)$. Then divide the numerator and denominator by $x^d$. If $p(x) = a_0 + a_1 x + \cdots + a_n x^n$ and $q(x) = b_0 + b_1 x + \cdots + b_m x^m$. This will yield an expression like $$ \frac{a_n x^{n-d} + a_{n-1} x^{n-1-d} + \cdots + a_0 x^{-d}}{b_m x^{m-d} + b_{m-1}x^{m-1-d} + \cdots + b_0 x^{-d}}. $$ All the exponents are nonpositive. In particular, only the exponents of the leading terms for each polynomial have a chance to be nonzero. So as $x \to \infty$, it's certain that every term except possibly for the leading terms go to $0$.
So if the degree of the numerator is bigger, then the limit is infinite. If the denominator is bigger, then the limit is $0$. And if they're the same, then the limit is $a_n/b_m$, which should be written as $a_d/b_d$ (since the degrees are the same).
Although written for polynomials, this methodology clearly applies for ratios of functions of the form $f(x) = \sum_{i = 0}^n a_i x^{d_i}$ for general real numbers $a_i$ and $d_i$.