I need some help to simplify the following: $$\frac{\sum_{k=0}^na_nr^n}{\sum_{k=0}^na_n(1+r)^n},$$ where $n \in \mathbb{N}$, $a_n$>0 and $r$>0.
More specifically, the question I have is $$\frac{\sum _{k=0}^{n-1} \frac{\Gamma(n+k)}{\Gamma(n-k)k!}\gamma ^{n-k}\left(\frac{1}{2}\right)^k}{\sum _{k=0}^{n-1} \frac{\Gamma(n+k)}{\Gamma(n-k)k!}(1+\gamma)^{n-k}\left(\frac{1}{2}\right)^k}.$$
Is there any other way to write into another form? Any suggestion and reference are welcome.
Rewrite $$\sum _{k=0}^{n-1} \frac{\Gamma(n+k)}{\Gamma(n-k)\,k!}\gamma ^{n-k}\left(\frac{1}{2}\right)^k=\gamma^n\sum _{k=0}^{n-1} \frac{\Gamma(n+k)}{\Gamma(n-k)\,k!}\left(\frac{1}{2\gamma}\right)^k=\gamma^n\sqrt{\frac{2\gamma}{\pi }}\, e^{\gamma }\, K_{\frac{1}{2}-n}(\gamma )$$ where appears the modified Bessel function of the second kind.
Just do the same for the denominator to get $$\frac{\sum _{k=0}^{n-1} \frac{\Gamma(n+k)}{\Gamma(n-k)k!}\gamma ^{n-k}\left(\frac{1}{2}\right)^k}{\sum _{k=0}^{n-1} \frac{\Gamma(n+k)}{\Gamma(n-k)k!}(1+\gamma)^{n-k}\left(\frac{1}{2}\right)^k}=\frac 1e\left(\frac{\gamma}{1+\gamma}\right)^{n+\frac 12}\frac{K_{\frac{1}{2}-n}(\gamma )}{K_{\frac{1}{2}-n}(\gamma +1)}$$