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?
 A: If $|x|>1$, let $y=\frac{1}{x}$. Then, $|y|<1$, and your series is 
$y+y^2+y^3+...$
This is a geometric series, for which the limit is known to be $\frac{y}{1-y}$, if $|y|<1$, and it diverges otherwise. Now, why is that?
Factorize $(1-y^{n+1})=(1-y)(1+y+y^2+...+y^n) \Rightarrow (1+y+y^2+...+y^n)=\frac{1-y^{n+1}}{1-y}$. 
If $|y|<1$, then $y^{n+1} \to 0$, and the sum is $(1+y+y^2+...+y^n)=\frac{1}{1-y}$. Subtracting $1$ on both sides gives the result. If $y=-1$, then the sum constantly oscilates between $+1$ and $-1$ (assuming these are real numbers), and if $y<-1$ it oscilates even more, so the series can't converge. If $y\geq 1$, then the series diverges because you're adding increasingly bigger numbers.
Now, your method is half right. It does give the sum of the series, provided there is one. The problem is that you are assuming there is such a number right of the bat, which could be incorrect: try plugging $a=+\infty$ to see that you method arrives at an indeterminate expression $\infty - \infty$
EDIT: Oh, and I forgot, but $\frac{y}{1-y} = \frac{\frac{1}{x}}{1-\frac{1}{x}}=\frac{1}{x-1}$, which is the sought result.
A: Your edit with the proof that $a = \frac{1}{x-1}$ is very good.  Some people mentioned that it only works if $\lvert x \rvert > 1$, and you indicated that you weren't sure about that, so maybe this might help:
Basically, what happens if $x = \frac{1}{2}$?
Then $\frac{1}{x} = \frac{1}{1/2} = 2$, so 
$$
\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} +\frac{1}{x^4}+ \ldots = 2 + 4 + 8 + 16 +\ldots
$$
It's probably clear that if you add all these numbers together, you should get infinity.  This is an example of a "divergent" series; the series isn't approaching a single number.
Anyway, when $x = \frac{1}{2}$, your formula predicts that
$$
2 + 4 + 8 + 16 + \ldots = \frac{1}{1/2 - 1} = \frac{1}{-1/2} = -2,
$$
which doesn't seem right.  So your formula $a = \frac{1}{x-1}$, which works very well when $x> 1$, does not seem to work for $x = \frac{1}{2}$.
This is not your fault!  Your formula is the "right" one.  But there are good reasons why your formula only works if $\lvert x \rvert > 1$.  (Some other things to try: what if $x  = 1$?  What if $x = -1$?  What if $x = 0$?  What if $x = -\frac{1}{2}$?)
To learn more, you might want to study Geometric Series:
http://en.wikipedia.org/wiki/Geometric_series
A: 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.
