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$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)}$$ 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 $\ln(\Gamma(1-n+ 1/a))$, or one of its variants that you can find at http://en.wikipedia.org/wiki/Stirling%27s_approximation

$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)}$$ If $a n < 1$, you'll want to use the reflection principle $$\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{(\pi z)}}$$ so $$ f(n) = \dfrac{a^{n-1} \Gamma(1/a)}{\Gamma(1-n+1/a)}$$

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

$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)}$$ Use Stirling's series for the asymptotic approximation of $\ln(\Gamma(n - 1/a))$, or one of its variants that you can find at http://en.wikipedia.org/wiki/Stirling%27s_approximation

$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)}$$ 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 $\ln(\Gamma(1-n+ 1/a))$, or one of its variants that you can find at http://en.wikipedia.org/wiki/Stirling%27s_approximation

$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)}$$ Use Stirling's series for the asymptotic approximation of $\ln(\Gamma(n - 1/a))$, or other variants that you can find at http://en.wikipedia.org/wiki/Stirling%27s_approximation

$$f(n) = \dfrac{(-a)^{n-1} \Gamma(n - 1/a)}{\Gamma(1 - 1/a)}$$ Use Stirling's series for the asymptotic approximation of $\ln(\Gamma(n - 1/a))$, or one of its variants that you can find at http://en.wikipedia.org/wiki/Stirling%27s_approximation