Find value of sum of reciprocals of powers of a number Is there a simple way to find the value of the following expression?
$$\frac1x+\frac1{x^2}+\frac1{x^3}+\cdots$$
On trial and error, I was getting $\frac1{x-1}$, but I'm looking for a mathematical proof to it.
Please don't use complicated notation like summation unless absolutely necessary, because I'm not too familiar with it.
Edit: I tried another method. Let the answer be $a$. If we calculate $ax$, we get what appears to be $1+a$.
$$ax=1+a$$
$$ax-a=1$$
$$a(x-1)=1$$
$$a=\frac1{x-1}$$
Is that sufficient to prove the answer?

Define the $n$th partial sum by
$$
S_n = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots + \frac{1}{x^n}
$$
Then
\begin{align*}
x S_n &= 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots + \frac{1}{x^{n-1}} \\
\implies x S_n - S_n &= 1 - \frac{1}{x^n} \iff S_n = \frac{1- \frac{1}{x^n}}{x-1}, \; x \neq 1
\end{align*}
Now assume $\left| \frac{1}{x} \right| < 1$. The limit $n \to \infty$ is then
\begin{align*}
S := \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1- \frac{1}{x^n}}{x-1} = \frac{1}{x-1},
\end{align*}
because $(1/x)^n$ goes to $0$ if $|1/x| < 1$.
If $|1/x| > 1$ the limit is $ \pm \infty$ and the sum diverges.
