Getting an improved upper bound for a finite geometric series The title says it all.
For an infinite geometric series, we know that if $|x| < 1$, then
$$\sum_{i=0}^{\infty}{x^i} = \frac{1}{1 - x}.$$
On the other hand, if $|x| \geq 1$, then the the infinite geometric series
$$\sum_{i=0}^{\infty}{x^i} = 1 + x + x^2 + x^3 + \cdots$$
diverges, so that we consider the case of a finite geometric series.
We know that the equation
$$Y = \sum_{j=0}^{k}{y^j} = 1 + y + \cdots + y^k = \frac{y^{k + 1} - 1}{y - 1}$$
holds.
Traditionally, an upper bound for $Y$ is obtained via
$$Y = \frac{y^{k + 1} - 1}{y - 1} < \frac{y^{k + 1}}{y - 1}$$
so that we get
$$Z = \sum_{j=0}^{k}{y^{-j}} = 1 + \frac{1}{y} + \cdots + \frac{1}{y^k} = \frac{Y}{y^k} < \frac{y}{y - 1}.$$

My question is this:  Is it possible to improve on the upper bound $$\frac{y}{y - 1}$$ for $$Z = \sum_{j=0}^{k}{y^{-j}}?$$  Or is this already the least upper bound for $Z$?

Of course, I would have to qualify that I am interested in upper bounds for $Z$ that do not depend on $k$.

If $y>1$,
$$\sum_{j=0}^{\infty}{y^{-j}}=\sum_{j=0}^{\infty}{\left(\frac 1y\right)^j}=\frac 1{1-1/y}=\frac{y}{y-1}$$
Since the terms are all positive, $\frac{y}{y-1}$ is not just the limit, it is the least upper bound to the sequence $\sum_{j=0}^{k}{y^{-j}}$. Therefore, for all limits that do not depend on $k$, that is the very best upper bound. No better bound is possible.
If $y<-1$, the series is an alternating series, and it alternates above and below your expression, therefore your expression is no longer an upper bound. The only upper bound that works for all $k$ is $1$, since that is the value of the sum for $k=0$ and all later sums are smaller. A bound that works for $k>0$ is $1+\frac 1y+\frac 1{y^2}$, and so on.
If $y=-1$, the series alternates between $1$ for even $k$ and $0$ for odd $k$. The least upper bound is $1$.
If $-1<y<0$ or $0<y\le 1$, your sequence diverges, and an upper bound that does not depend on $k$ is impossible.
If $y=0$, the sum is undefined.
