Another way to find $\sum_{n=1}^{\infty}{\dfrac1{k^n}}$ Find sum
$$S=\sum_{n=1}^{\infty}{\dfrac1{k^n}}$$
where $k\in\mathbb{Z^+}\setminus\{1\}$.
In my book they used limits to find this sum. They wrote
$$S=-1+\sum_{n=0}^{\infty}{\dfrac1{k^n}}$$
and then
$$S=-1+\lim_{n\to\infty}{\dfrac{\dfrac{1}{k^{n+1}}-1}{\dfrac1k-1}}$$
Calculating this limit, they got
$$S=\dfrac1{k-1}$$
This solution is very complicated and I tried to find $S$ in different way. I wrote $S$ in number base $k$:
$$S=0.1_k+0.01_k+0.001_k+\dots=0.\overline1_k$$
Then I had
$$kS=1.\overline1_k$$
$$kS-S=1.\overline1_k-S$$
$$(k-1)S=1$$
and finally
$$S=\dfrac{1}{k-1}$$
Is my solution mathematically correct and acceptable?
 A: Here is another way of expressing your argument. Suppose that
$$ S = \sum_{n=1}^\infty \frac{1}{k^n}. $$
Then
$$ kS = \sum_{n=1}^\infty \frac{1}{k^{n-1}} = \sum_{n=0}^\infty \frac{1}{k^n} = 1 + S. $$
Therefore $(k-1)S = 1$ and $S = 1/(k-1)$.
What are the problems in this argument, if any? You need to be careful in what it means that $S$ is the sum of the original series, and whether it implies that your reasoning is valid. Our reasoning involved two steps. Step 1 was
$$ S = \sum_{n=1}^\infty a_n \Longrightarrow cS = \sum_{n=1}^\infty ca_n. $$
Step 2 was
$$ \sum_{n=1}^\infty a_n = a_1 + \sum_{n=1}^\infty a_{n+1}. $$
These two steps might seem obvious, but they are features of the non-trivial definition of the sum of an infinite series. While in this case nothing surprising happens and you can actually prove these two formulas whenever the sums involve converge, in other cases (mostly involving conditionally convergent series) intuition could be misleading. You just need to be careful.
As an aside, note that while the sum only converges when $|k|>1$, the formula makes sense for all $k \neq 1$. It's not clear what to make of the computation, however, and how to interpret the result. This is actually possible using the theory of analytic continuation, which perhaps you'll get to learn in the future.
A: Here's the standard way
to make your proof rigorous:
Let
$S_m$
be the sum of the 
first $m$ terms,
so
$S_m
= \sum_{n=1}^{m}{\dfrac1{k^n}}
$.
Then
$\begin{array}\\
k S_m
&=k\sum_{n=1}^{m}{\dfrac1{k^n}}\\
&=\sum_{n=1}^{m}{\dfrac1{k^{n-1}}}\\
&=\sum_{n=0}^{m-1}{\dfrac1{k^{n}}}\\
&=1+\sum_{n=1}^{m-1}{\dfrac1{k^{n}}}\\
&=1+\sum_{n=1}^{m}{\dfrac1{k^{n}}}-\dfrac1{k^m}\\
&=1-\dfrac1{k^m}+S_m\\
\end{array}
$
so
$S_m(k-1)
=1-\dfrac1{k^m}
$
or
$S_m
=\dfrac{1-\dfrac1{k^m}}{k-1}
=\dfrac{1}{k-1}-\dfrac{1}{k^m(k-1)}
$.
If $k > 1$,
then 
$\frac1{k^m}
\to 0$
as $m \to \infty$,
so
$S_m
\to\dfrac{1}{k-1}
$
as $m \to \infty$.
Here is a proof 
(also not original)
that
$1/k^m \to 0$:
Since $k > 1$,
$k = 1+c$
where $c > 0$.
By Bernoulli's inequality
(easily proved by induction),
$k^m
=(1+c)^m
> 1+cm
> cm
$,
so
$\frac1{k^m}
<\frac1{cm}
<\frac1{(k-1)m}
$.
