Summation with absolute value $\sum_{k=-2}^{\infty}(\frac{1}{2})^{k}\alpha^{|n-k|}$

Question

How can I calculate explicitly $\sum_{k=-2}^{\infty}(\frac{1}{2})^{k}\alpha^{|n-k|}$ for every whole number n? Note that $-1<\alpha<1$

Thank you

Answer 1

Hint: $|n-k|=n-k$ if $k<n$, and $|n-k|=k-n$ if $k\ge n$. So $$\sum_{k=-2}^{\infty}\Bigl(\frac{1}{2}\Bigr)^{k}\alpha^{|n-k|} =\sum_{k=-2}^{n-1}\Bigl(\frac{1}{2}\Bigr)^{k}\alpha^{n-k} +\sum_{k=n}^{\infty}\Bigl(\frac{1}{2}\Bigr)^{k}\alpha^{k-n}$$ which is a sum of two GPs. You will have to be careful with the first one if $\alpha=\frac{1}{2}$.