Geometric explanation for the convergence of $\sum_{n = 1}^{\infty} (1/a)^n$ when $1I'm studying some series and I'm trying to see them in a more geometric way, thinking of integrals or parts of something big (you will know what i mean). But I can't find a geometric way to see why those should converge!
Being a bit more precise, what I want is:

Let $S = \sum_{n = 1}^{\infty} (1/a)^n$ be a series with $a > 1$. Then, $S < \infty$.

If $a = 2$, using the "part of a hole" idea works, because we can think of the circle with radius $1$ and then the sum $1/2 + 1/4 + \cdots + 1/\infty$ is just filling this circle by painting half of it, then taking the rest and painting half of it, doing this recursively.
Then, if $a > 2$, obviously the series will converge because $(1/a)^n \leq (1/2)^n$ and so the sum will never get to fill the circle, hence it must be finite.

Now, the problem lies in $2 > a > 1$. Is there a geometric argument that is somehow intuitive for why this should converge?

 A: Note that $$\sum_{n=1}^\infty \left(\frac{1}{a} \right)^n=\frac{1}{a-1}$$
For $a=2$, the sum converges to $1$, so we can easily visualize the sum as a progressive filling of fractions of a circle. For $a>2$, the evidence that $(1/a)^n<(1/2)^n$  also allows an intuitive visualization of the convergence to a limit $<1$.
For $1<a<2$, since the sum converges to a number that is $>1$, to keep the possibility of easy visualization of the convergence, it is sufficient to rescale the fractions filled in the circle by a factor $a-1$. In this way, our limit is again $1$ as in the case $a=2$.
For example, set $a=3/2$. We know that the sum converges to $1/(3/2-1)=2$. So we must rescale our fractions by a factor $2$. Take the first term of the sum $1/(3/2)=2/3$, divide it by $2$, and we get $1/3$: so we fill $1/3$ of the circle and $ 2/3$ of it remain to be filled. Take the second term of the sum $1/(3/2)^2=4/9$, divide it by $2$, and we get $2/9$: so we fill other $2/9$ of the circle, arriving to a total filled portion of $5/9$ of the circle, and $ 4/9$ of it remain to be filled. Continuing in this way, the proportion that remains to to be filled is $8/27$ in the third term, $8/81$ in the fourth, and so on.
In general, after summing $m$ terms for a given $a$, the proportion remaining to be filled is $$\left(\frac{1}{a}\right)^m$$
which can be easily visualized from a geometric point of view.
A: Consider the following continuous function:
$$f(s) = \frac{1}{a^s}$$
Compare this to the left-continuous step function:
$$g(s) = \frac{1}{a^{\lceil s \rceil}}$$
Note that
$$\int_0^n g(s) ds \equiv \sum_{i=1}^{n} \frac{1}{a^i} \quad \forall n \in \mathbb{N}$$
Also note that $\lceil s \rceil \geq s \implies f(s) \geq g(s)\quad\forall s \in \mathbb{R}$.
Therefore, we know
$$\int_0^n f(s) ds \geq \int_0^n g(s) ds \quad \forall n\geq0$$
Geometrically, we can see that the area under the curve for $f(s)$ includes the area under the curve $g(s)$ as a subregion.
We can evaluate the improper integral of $f(s)$ to get a closed-form result:
$$\lim_{b\to\infty}\int_0^b \frac{1}{a^s} ds =\lim_{b\to\infty}\frac{1}{\ln(a)}\left[ 1- a^{-b}\right] = \frac{1}{\ln(a)} \implies \lim_{b\to\infty}\int_0^b g(s) ds < \infty $$
$$\implies \sum_{n=1}^{\infty} \frac{1}{a^n} < \infty$$
