Upper bounds on the partial sums of a sequence of the from $a_{i+1} = \alpha^{n-i}a_i$. Let $\alpha <1, k\le n$. Consider the sequence $$a_0 = 1, a_{i+1} = \alpha^{n-i} a_i.$$
I'm trying to upper bound the sum $$S(\alpha, n, k) := \sum_{i = 0}^k a_i.$$  Note that the sequence $\{a_i\}$ is dominated by a geometric sequence with the ratio $\alpha^{n-k}$, but the upper bound that results from this is too weak for the problem I want to apply this to. I also tried splitting the sum into blocks and separately bounding each block by a geometric series, but this quickly turned intimidatingly messy. Is there a trick I can use to exploit the stronger initial decay? Alternately, is there a neat way to deal with the multiple geometric series  approximation? 
In case helpful, it may be assumed that $k\gg 1$, and that $\alpha \ll 1$.
 A: First, we can get an explicit formula for $a_i$ with $i \in N$ using the recursive definition.
$$a_i=\prod_{m=0}^{i-1}{\alpha^{n-m}}=\alpha^{\sum_{m=0}^{i-1}{(n-m)}}=\alpha^{\sum_{m=0}^{i-1}{(n)}-\sum_{m=0}^{i-1}{(m)}}$$
$$a_i=\alpha^{ni-\frac{i(i-1)}{2}}=\alpha^{ni+\frac{i}{2}-\frac{i^2}{2}}=\alpha^{\frac{ni+i}{2}+\frac{ni-i^2}{2}}$$
$$\sum_{i=0}^{k}{a_i}=\sum_{i=0}^{k}{\alpha^{\frac{ni+i}{2}+\frac{ni-i^2}{2}}}$$
In the summation $n \geq k \geq i$, which implies that $ni \geq i^2$ and $\frac{ni-i^2}{2} \geq 0$.
$$\frac{ni+i}{2} \leq \frac{ni+i}{2}+\frac{ni-i^2}{2}$$
I suggest to add a lower bound on $\alpha$ such that $0 < \alpha < 1$
$$\alpha^{\frac{ni+i}{2}} \geq \alpha^{\frac{ni+i}{2}+\frac{ni-i^2}{2}}$$
$$\sum_{i=0}^{k}{\alpha^{\frac{ni+i}{2}}} \geq \sum_{i=0}^{k}{a_i}$$
$$\sum_{i=0}^{k}{\alpha^{\frac{ni+i}{2}}}=\sum_{i=0}^{k}{\left(\alpha^{\frac{n+1}{2}}\right)^i}=\sum_{i=0}^{k}{\left(\sqrt{\alpha}^{(n+1)}\right)^i}$$
This is in the form of a partial geometric sum:
$$\frac{1-\sqrt{\alpha}^{(n+1)(k+1)}}{1-\sqrt{\alpha}^{(n+1)}}$$
Conclusion:
$$S(\alpha, n, k) \leq \frac{1-\sqrt{\alpha}^{(n+1)(k+1)}}{1-\sqrt{\alpha}^{(n+1)}}$$
$$0<\alpha<1 \quad\quad 1 \leq k \leq n$$
This approximation tends to work best for large $n$, small $\alpha$, and $n \gg k$.
