Show $\lim_{n\to\infty}\frac{\sum_{k=1}^{n}(\frac{\Gamma(k-c\lambda)}{\Gamma(k+a\lambda)})^{2}}{n^{1-2(a+c)\lambda}}=\frac{1}{1-2(a+c)\lambda}$

Question

I'm trying to compute:$$\lim_{n\to\infty}\frac{\sum_{k=1}^{n}(\frac{\Gamma(k-c\lambda)}{\Gamma(k+a\lambda)})^{2}}{n^{1-2(a+c)\lambda}}=\frac{1}{1-2(a+c)\lambda}$$ where $-1<a<\frac{1-c}{2}$, $\lambda=\frac{1}{1+c}$.

I tried using $$ \lim _{n \rightarrow \infty} \frac{\Gamma(n+\alpha)}{\Gamma(n) n^{\alpha}}=1, \alpha\in \mathbb{R}$$ but it didn't work out.

I'm quite sure of my result but I could not check it.

Thank you very much!