I've been looking into the following sum for values of x:

$$\sum_{n=0}^{\infty} {\frac{1}{x ^ n}}$$

And after plugging in different values of x, I became confident enough to make the following conjecture:

The limit of the expression, $\sum_{n=0}^{\infty} {\frac{1}{x ^ n}}$ , will be equal to ${\frac{x}{x-1}}$ for any value of x greater than one.

My question:

Can someone provide a proof or explanation to this conjecture? And if so, what is it?

If unable or unwilling to provide a proof, suggestions on how to prove my conjecture would also be satisfactory, though be ready to clarify and elaborate some of your points.

I attempted to prove the conjecture via a contradiction proof by seeing what would result if this conjecture was false (then search for a contradiction thus making my conjecture true). However, I couldn't extrapolate any resulting truths which would occur if my conjecture was false. I tried rewriting the sum as various forms such as, $\sum_{n=0}^{\infty} {\frac{x}{x ^ {n+1}}}$ and $\sum_{n=0}^{\infty} {x^{-n}}$ to help, yet I was unable to get any closer to proving my conjecture.

Important note: My formal education of mathematics only goes up to Algebra 1 in high school, so if simpler notation and mathematic vocabulary can be used in an answer it would be greatly appreciated. Thank you.


2 Answers 2


This is what is known as a geometric series. Start with some $r$ with $r\neq 1$ (for simplicity): $$ S_n=1 + r + r^2 + \cdots +r^n $$Note that if we multiply by $(1-r)$ we have $$ (1-r)S_n = (1 + r + r^2 + \cdots +r^n)-( r + r^2 + \cdots +r^n+r^{n-1}); $$then all the intermediate terms cancel: $$ (1-r) S_n = 1 - r^{n+1};\qquad S_n = \frac{1-r^{n+1}}{1-r} $$For example, if $r=3,n=4$ we have $S_n = 1+3+3^2+3^3+3^4=121 = \frac{1-243}{1-3}$. Now if $|r|<1$, we can take the limit and obtain the infinite sum: $$ \sum_{r=0}^{\infty}r^n = \lim_{n\to\infty} S_n = \lim_{n\to\infty}\frac{1-r^{n+1}}{1-r}=\frac{1}{1-r}$$ For example, if $r=1/2$, we have $1+1/2+1/4+1/8+1/16+\cdots =2 = 1/(1-1/2)$. Try drawing some pictures to see this.

Now for your question: if $|x|>1$, $|x|^{-1}<1$. So we have $$ \sum_{r=0}^{\infty}\frac{1}{x^n} = \sum_{r=0}^{\infty}\left(\frac{1}{x} \right)^n = \frac{1}{1-1/x}=\frac{x}{x-1} $$

  • $\begingroup$ There are three things I'm not able to follow: How do you simplify ${\frac{1 - r^{n+1}}{1 - r} to {\frac{1}{1 - r}}? Also, how do the last two simplifications work on the bottom? $\endgroup$
    – Tauist
    Jun 11, 2020 at 4:23
  • 1
    $\begingroup$ 1) If $|r|<1$, $\lim_{n\to\infty}r^n=0$. 2) If $|r|<1$, $\sum_{n=0}^{\infty} r^n = 1/(1-r)$ 3) $1/x^n = (1/x)^n$ 4) $1/(1-1/x)\cdot x/x = x/(x-1)$ $\endgroup$
    – Integrand
    Jun 11, 2020 at 4:45
  • $\begingroup$ I understand this now! One thing I'm curious on (if you're aware): where did the idea of multiplying the sum by (1 - r) come from? It seems arbitrary and random, I would have never thought to do that while trying to prove my statement. $\endgroup$
    – Tauist
    Jun 11, 2020 at 6:59

You can prove, by induction or whatever, that for any $\;n\in\Bbb N\;$ we have

$$1+x+x^2+\ldots+x^n=\sum_{k=0}^n x^n=\frac{1-x^{n+1}}{1-x}$$

The above is just a geometric sequence, and to get the infinite sum (series), we must take the limit when $\;n\to\infty\;$ of the above, which is nothing but the n-th term of the sequence of partial sums of the geometric series, and thus

$$\sum_{k=0}^\infty x^k=\lim_{n\to\infty}\sum_{k=0}^n x^k=\lim_{n\to\infty}\frac{1-x^{n+1}}{1-x}=\begin{cases}\text{doesn't exist finitely},&|x|\ge1\\{}\\\frac1{1-x},&|x<1\end{cases}$$

Well, now in your case, do the same with the infinite geometric series of $\;\sum\limits_{k=0}^\infty\left(\cfrac1x\right)^k\;$, which converges iff $\;\left|\cfrac1x\right|<1\iff |x|>1\;$ ...and then substitute.


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