Closed form expression for $\sum_{k=0}^{n} \alpha^{-k}$?
\begin{align}
(1-x)\sum_{k=0}^{n}x^k &amp;= \sum_{k=0}^{n}x^k - x\sum_{k=0}^{n}x^k \\
\\
                       &amp;= \sum_{k=0}^{n}x^k - \sum_{k=1}^{n+1}x^k \\
\\
&amp;=1 + x+x^2+ \dots+x^n \ \\
&amp; \qquad -x-x^2-\dots-x^n-x^{n+1}\\
\\
&amp;= 1 - x^{n+1}
\end{align}

Thus if $x \ne 1$, dividing both sides by $(1-x)$ results in
$$\sum_{k=0}^{n}x^k = \frac{1-x^{n+1}}{1-x}$$

Replace $x = \alpha^{-1}$, it becomes
$$\sum_{k=0}^{n}\alpha^{-k} = \frac{1-(1/\alpha)^{n+1}}{1-1/\alpha}$$