I would appreciate any feedback about the correctness of the proof and some potential other techniques! Proposition: Let $m, n \in \mathbb{R}$, such that $\vert n \vert \gt 1.$ Then the series $\displaystyle \sum_{k=1}^{\infty}\frac{m}{n^k}$ converges to $\dfrac{m}{n-1}$.
Proof: We observe, that this is a geometric series in the form of $m\times n^{-k}$. Since the series is a geometric series, which converges, (EDIT) because if $\vert n \vert \gt 1$, then $\vert \frac{1}{n}\vert \lt 1$. Thus, we can use the geometric series test to first, find the common ration and then compute the sum. The common ratio, $r=\dfrac{a_{n+1}}{a_n}$, after substituting $a_n$, we have: $\frac{m}{n^2}:\frac{m}{n}$ $\implies \frac{mn}{n^2 m}$. After simplifying the expression, we get $r=\frac{1}{n}$. Therefore, we can evaluate the sum of the series using the closed form, $S_n=\frac{a_1}{1-r}$. Substituting $r$, we get: $S_n=\dfrac{a_1}{1-\frac{1}{n}}$. After doing the algebraic manipulations, we get $S_n=\dfrac{m}{n-1}$, as desired.
Another way is to look at the partial sums and use $\dfrac{1-x^j}{1-x} =\sum_{k=0}^{j-1} x^k $.
$\begin{array}\\ \sum_{k=1}^{j-1}\dfrac{m}{n^k} &=\frac{m}{n}\sum_{k=1}^{j}\frac{1}{n^{k-1}}\\ &=\dfrac{m}{n}\sum_{k=0}^{j-1}\frac{1}{n^{k}}\\ &=\dfrac{m}{n}\dfrac{1-\frac1{n^j}}{1-\frac1{n}}\\ &=m\dfrac{1-\dfrac1{n^j}}{n-1}\\ &=\dfrac{m}{n-1}-\dfrac{m}{n^j(n-1)}\\ \end{array} $
so $\left|\sum_{k=1}^{j-1}\dfrac{m}{n^k}-\dfrac{m}{n-1}\right| =\dfrac{m}{n^j(n-1)} $.
Therefore, to make $\left|\sum_{k=1}^{j-1}\dfrac{m}{n^k}-\dfrac{m}{n-1}\right| \lt \delta $, is is enough to make $\dfrac{m}{n^j(n-1)} \lt \delta$, and this can be done by making $j$ large enough so that $n^j > \dfrac{m}{\delta(n-1)} $.
If you are allowed the use of logs, this is just $j \gt \dfrac{\log(m)-\log(\delta(n-1))}{\log(n)} $.
If not, you can use Bernoulli's inequality ($(1+x)^j \ge 1+xj > xj$) to have $n^j = (1+(n-1))^j \gt (n-1)j $ so $\dfrac{m}{n^j(n-1)} \lt \dfrac{m}{j(n-1)(n-1)} = \dfrac{m}{j(n-1)^2} $. Therefore, if $\dfrac{m}{j(n-1)^2} \lt \delta$, or $j > \dfrac{m}{\delta(n-1)^2} $, we have $\dfrac{m}{n^j(n-1)} \lt \delta$.
This is, of course, much worse than the inequality with log, but it is completely elementary.