$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)}$$ If $a n < 1$, you'll want to use the reflection principle $$\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{(\pi z)}}$$ so $$f(n) = \dfrac{(a)^n \pi}{\sin(\pi/a) \Gamma(1-1/a) dfrac{a^{n-1} \Gamma(1-n+1/a)}$$Gamma(1/a)}{\Gamma(1-n+1/a)}$$Use Stirling's series for the asymptotic approximation of \ln(\Gamma(1-n+ 1/a)), or one of its variants that you can find at http://en.wikipedia.org/wiki/Stirling%27s_approximation 3 added 201 characters in body$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)}$$If a n < 1, you'll want to use the reflection principle$$\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{(\pi z)}}$$so$$ f(n) = \dfrac{(a)^n \pi}{\sin(\pi/a) \Gamma(1-1/a) \Gamma(1-n+1/a)}$$Use Stirling's series for the asymptotic approximation of \ln(\Gamma(n - \ln(\Gamma(1-n+ 1/a)), or one of its variants that you can find at http://en.wikipedia.org/wiki/Stirling%27s_approximation 2 added 5 characters in body$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)} Use Stirling's series for the asymptotic approximation of $\ln(\Gamma(n - 1/a))$, or other one of its variants that you can find at http://en.wikipedia.org/wiki/Stirling%27s_approximation