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?


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:
$$\lim_{b\to\infty}\int_0^b \frac{1}{a^s} ds =\lim_{b\to\infty}\frac{1}{\ln(a)}\left[ 1- a^{-b}\right] = \frac{1}{\ln(a)} \implies \lim_{b\to\infty}\int_0^b g(s) ds < \infty $$
$$\implies \sum_{n=1}^{\infty} \frac{1}{a^n} < \infty$$
