TITLE: Find value of sum of reciprocals of powers of a number
QUESTION [9 upvotes]: 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?

REPLY [4 votes]: 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