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?

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    $\begingroup$ This is a geometric series, consecutive terms always have the ratio $\frac{1}{x}$, so yes, there is a simple expression for the value (and a simple way to prove it). What have you tried? $\endgroup$
    – hardmath
    Commented Dec 7, 2014 at 14:57
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    $\begingroup$ Probably you should mention that $x>1$ because otherwise the sum diverge. $\endgroup$
    – user170039
    Commented Dec 7, 2014 at 14:58
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    $\begingroup$ Remarking on Edit: What you write is sufficient to show if there is "the answer" $a$, then $a$ must be $\frac{1}{x-1}$. Note what others have said about divergence when $|x| \le 1$. $\endgroup$
    – hardmath
    Commented Dec 7, 2014 at 15:04
  • $\begingroup$ What do you mean by divergence? how can there not be an answer? $\endgroup$ Commented Dec 7, 2014 at 16:17
  • $\begingroup$ For example, if $x = 0.5$. Then $a$ would be infinite. Another example is $x = -1$, then there is no $a$ (the partial sums are $-1,0,-1,0,...$). In both cases we say that the series diverge. $\endgroup$
    – Wood
    Commented Dec 8, 2014 at 0:42

3 Answers 3


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.

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    $\begingroup$ I thought he said he didn't want complicated notation. $\endgroup$
    – tcrosley
    Commented Dec 7, 2014 at 22:26

If $|x|>1$, let $y=\frac{1}{x}$. Then, $|y|<1$, and your series is


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.

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    $\begingroup$ I'm not sure this really would help the OP to prove the "trial and error" answer proposed unless the results for geometric series were already known to them. Perhaps, if you insist on posting an Answer, you should give a thorough one. $\endgroup$
    – hardmath
    Commented Dec 7, 2014 at 15:00
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    $\begingroup$ Sorry, I didn't get what you mean. $\endgroup$ Commented Dec 7, 2014 at 15:02

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:


  • $\begingroup$ Thank you, your answer was very helpful. $\endgroup$ Commented Dec 8, 2014 at 10:14

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