This is not a homework assignment, but a problem that is bothering me nonetheless. I feel like I should be able to figure this out, but either I'm missing some detail, or I'm doing something completely wrong.

The question arises because I read a theorem in my elementary number theory book that states the following:


and it goes on to prove it using induction. This got me thinking, because I recall from my Calculus class over the summer that the power series $\sum_{i=0}^n x^i=\frac{1}{1-x}$. So I just decided to try and see if this is right by taking my knowledge from Calculus and using a bit of algebra to manipulate the fraction.

So what I come up with is the following:

\begin{align} \sum_{i=0}^{n-1} x^i&=\sum_{i=0}^{n} x^i-x^n\\ &=\frac{1}{1-x}-x^n\\ &=\frac{1-x^n+x^{n+1}}{1-x}\\ &=\frac{x^n-1-x^{n+1}}{x-1} \end{align}

I'm having trouble figuring out how to deal with the $-x^{n+1}$ in the numerator in order to make the fraction equal $\frac{x^n-1}{x-1}$. Where am I going wrong?


We do the calculation in the style you are trying to use, and then explain why one should not do that. We have $$1+x+x^2+\cdots+x^{n-1}=(1+x+x^2+x^3+\cdots)-x^n(1+x+x^2+x^3+\cdots).$$ Using the familiar formula for the sum of an infinite geometric series, we get $$1+x+x^2+\cdots+x^{n-1}=\frac{1}{1-x}-x^n\frac{1}{1-x},$$ which readily simplifies to the form you want.

Caveat: The formula $1+x+x^2+x^3+\cdots =\frac{1}{1-x}$ holds only when $|x|\lt 1$. So if we are going to use it for our finite geometric series, we will only have shown that the desired formula holds when $|x|\lt 1$. In fact, the formula for the sum of a finite geometric series holds for all $x\ne 1$.

Another reason to eschew the infinite series argument is that the most standard proofs for the sum of an infinite geometric series use the formula for the sum of a finite geometric series.

  • 1
    $\begingroup$ Indeed, I had forgotten that the formula $\frac{1}{1-x}$ only applies when summing to infinite. Had I considered subtracting $x^n(\sum_{i=0}^{\infty} x^i)$ I would have gotten it. $\endgroup$ – Mirrana Oct 12 '13 at 3:07

In order to show that $\frac{x^n-1}{x-1}=1+x+...+x^n$ just multiply by $x-1$ both sides and observe that most terms cancel.

The calculus result states that $\sum_{n=0}^\infty x^n=\frac{1}{x-1}$ for $|x|<1$, this follows from the previous result by noting that $\lim_{n\to\infty} x^n=0$. You seem to be confusing the series with the partial sum.


Hint: $$ A=1+x+x^2+x^3+\cdots+x^{n-1}\\ xA=x+x^2+x^3+\cdots+x^{n}\\ \implies xA-A=x^n-1 $$

  • $\begingroup$ That is indeed another good proof that I hadn't considered. I just wanted to see if my idea worked, and couldn't figure out what I was doing wrong. $\endgroup$ – Mirrana Oct 12 '13 at 3:11
  • $\begingroup$ I like your proof. Very elegant. $\endgroup$ – Umberto Oct 12 '13 at 6:08

I think the difficulty here is that you have confused an infinite geometric series that converges with a finite one.

There is, indeed, a formula:

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

where $|x| < 1$.

But you have used this formula for a finite geometric series.

Then it makes no sense.

(For example, $1/2 + 1/4 + 1/8 \neq 1$; you would need to continue with $1/16 + 1/32 + \ldots$)


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