I want to prove that $\sum_{k=0}^\infty ar^k = \frac{a}{1-r}$ when $|r| < 1$. If I can prove the sequence of partial sums $S_n$ converges to said formula, then I will have proven the series does as well. To derive a formula for $S_n$ I considered the following:
$S_n = a + ar +ar^2 + \cdots + ar^{n-1}$,
$rS_n = ar + ar^2 + ar^3 + \cdots + ar^n$
$S_n - rS_n = a - ar^n \rightarrow S_n = \frac{a(1-r^n)}{1-r}$
I now want to show $\{S_n \}_{n=1}^\infty \rightarrow \frac{a}{1-r}$. By the algebraic limit theorem, $\lim_{n \to \infty} \frac{a(1-r^n)}{1-r} = \left(\frac{a}{1-r}\right) \lim_{n \to \infty} (1-r^n)$. Thus we must show that $\lim_{n \to \infty} (1-r^n) \rightarrow 1$. Further simplification with the algebraic limit theorem reduces this to showing that $\lim_{n \to \infty} r^n \rightarrow 0$.
Here is where I am unsure of my approach. By inspection one can easily tell that this limit goes to zero, but in the pursuit of rigor I wanted to prove it directly. To start, I will ignore the trivial case when $r = 0$. We can prove inductively that $\{r^n\}_{n=1}^\infty$ is monotonically decreasing or increasing in either case that $r$ is positive or negative respectively. Observe that by this monotonicity of the sequence and the assumption that $|r| < 1$ we can say that $\{r^n\}_{n=1}^\infty$ is bounded. Since the sequence is bounded, we can say it converges to some limit point $L$ by the Monotone Convergence Theorem.
My idea was that since we know this sequence converges, we could try to do some algebra to prove $L = 0$. Since $\{r^n\}_{n=1}^\infty$ converges, we know that $\{r^{n+1}\}_{n=1}^\infty \rightarrow L$ since this is the same sequence simply shifted a term. Now:
$r^{n+1} = r^n \cdot r$ and $\lim_{n \to \infty} r^{n+1} = \lim_{n \to \infty} r^n \cdot r \rightarrow L = Lr$ ....................................................(1)
Such an expression is only true if $r = 1$ or $L = 0$, yet $r=1$ contradicts our assumption that $|r| < 1$ so thus $L=0$.
It then follows that $\lim_{n \to \infty} \frac{a(1-r^n)}{1-r} = \left(\frac{a}{1-r}\right) \lim_{n \to \infty} (1-r^n) = \frac{a}{1-r}$ by the algebraic limit theorem.
My question is, is it allowed to do this sort of manipulation in $(1)$ and the logic that precedes it? I've seen similar manipulations done before, but it was in different contexts.