We know that the formula for computing a geometric series is:$$\sum_{i=1}^{\infty}{a_0r^{i-1}} = \frac{a_0}{1-r}$$ Out of curiosity, I would like ask: Is there any ways the formula can be derived other than the following two ways?

Method 1 (The way I found on my own): $$\sum_{i=1}^{\infty}{a_0r^{i-1}} \equiv S$$ $$S = a_0r^0+a_0r^1+a_0r^2+\cdots$$ $$S = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots\right)$$ $$S = r\left(a_0r^{-1} + S\right)$$ $$S = a_0 + rS$$ $$(1-r)S = a_0$$ $$S = \frac{a_0}{(1-r)}$$ Note that for this to work, you must first confirm this: $$\lim_{n\to\infty} a_n = 0$$

Method 2 (The way I found on the web): $$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$ $$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}$$ $$rS_n = r\left(a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}\right)$$ $$rS_n = a_0r^1 + a_0r^2 + a_0r^3 + \cdots + a_0 r^{n-1} + a_0 r^{n}$$ $$S_n-rS_n = a_0r^0 - a_0r^n$$ $$(1-r)S_n = a_0 - a_0 r^n$$ $$S_n = \frac{a_0(1 - r^n)}{1-r}$$ Given:$$\left|r\right| < 1,$$ $$\lim_{n\to \infty} S_n = \lim_{n\to \infty}\frac{a_0(1 - r^n)}{1-r} = \frac{a_0}{1-r}$$

I personally prefer Method 1 because it is faster and more intuitive, as we don't have to multiply by $r$.

Method 1 for formula of partial sums: $$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$ $$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots+a_0r^{n-2}+a_0r^{n-1}$$ $$S_n = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots+a_0r^{n-3}+a_0r^{n-2}\right)$$ $$S_n = r\left(a_0r^{-1} + S_n - a_0r^{n-1}\right)$$ $$S_n = a_0 + rS_n - a_0r^{n}$$ $$(1-r)S_n = a_0 - a_0r^n$$ $$S_n = \frac{a_0(1 - r^n)}{(1-r)}$$

  • $\begingroup$ The two are effectively equivalent but the second method views the infinite series as a sequence of partial sums, which is more amenable to proofs and is more rigorous. I'm not sure if there are other ways to prove it. Maybe there is a way with what are known as Fourier series, as a lot of series can be stumbled upon in that way, but it's not that instructive. $\endgroup$ Apr 23, 2013 at 18:45
  • $\begingroup$ The limit of the partial sums is the more rigorous way. You have to worry about convergence of the infinite sums to begin with otherwise. And doing it that way, you get an intermediate formula for the partial sum. $\endgroup$
    – rajb245
    Apr 23, 2013 at 18:47
  • $\begingroup$ You could also find the Taylor series for $\frac1{1-x}$, it's not hard to get a formula for the $n$-th derivative by induction. $\endgroup$
    – Javier
    Apr 23, 2013 at 18:52
  • 2
    $\begingroup$ The $r^{-1}$ in your first example makes it ugly. The simple variation to your first example is to note that $S=a_0 + rS$. $\endgroup$ Apr 23, 2013 at 19:13
  • 1
    $\begingroup$ Here's one way. $\endgroup$ Apr 23, 2013 at 19:24

5 Answers 5


If by derive, you mean go from the summation to the fraction representation, you probably identified the best ways of doing it. However, here's one non-rigorous way to get the result going the other way, i.e., starting with the fraction

$$ \frac{a_0}{1-r} $$

Some have observed that you can write the Taylor series for that at $r=0$. Another way is to use synthetic division or polynomial long division. It's hard to typeset here, but I'll give you the flavor as best I can. Think of the long division algorithm we learned in grade school, where you are generating the terms on the top one at a time as you are dividing the dividend by the term $1-r$, multiplying the newly generated term by the divisor, subtracting, and iterating:

$$ \begin{matrix} {} & a_0 & +a_0 r & +a_0 r^2 & +a_0 r^3 &+\cdots\\\hline 1-r)&a_0\\ {}&-a_0&+a_0 r\\\hline {}&{}&a_0 r\\ {}&{}&-a_0 r & +a_0 r^2\\\hline {}&{}&{}&a_0 r^2\\ {}&{}&{}&-a_0 r^2&+a_0 r^3\\\hline {}&{}&{}&{}&a_0 r^3\\ {}&{}&{}&{}&-a_0 r^3&+a_0 r^4\\\hline {}&{}&{}&\vdots&{} \end{matrix} $$

The process never terminates, but does successively give additional terms of the expansion you are asking about. After conjecturing the series generated represents the function, you of course have to check convergence and prove the formula's correctness, but it works out in this case.

  • $\begingroup$ Yes, I know what synthetic division is. This looks good. $\endgroup$
    – Justin
    Apr 23, 2013 at 22:26
  • $\begingroup$ Glad you liked it. I couldn't typeset it how I write it on paper, so I tried to explain a bit too much in the text :-) $\endgroup$
    – rajb245
    Apr 23, 2013 at 22:39











Expand the right hand side as a Taylor series around 0. The $n$th derivative is $a_0 n!$ at zero which gives the result. (You'd still have to verify convergence though, so it's not very rigorous at all.)

Ultimately, anything rigorous has to deal with the limit of partial sums on the left, so don't expect much variety in analysis type arguments.


Define $S_n=a_0+a_0r+a_0r^2+\cdots+a_0r^n$, then we have $$S_n=\frac{a_0(1-r^{n+1})}{1-r}$$ Now taking limit as $n$ tends to infinity, the result follows. To prove the formula for $S_n$, we consider $(1-r)(1+r+r^2+\cdots+r^n)=(1+r+r^2+\cdots+r^n)-(r+r^2+\cdots+r^{n+1})=1-r^{n+1}$. Multiplying both sides by $a_0(1-r)^{-1}$, we are done.

Alternatively, you can use induction to get the formula for $S_n$ , suppose $1+r+\cdots+r^{n−1} =(1−r^n )/(1−r)$ ,then for $n$ , $1+r+\cdots+r^{n−1}+r^n =(1−r^n )/(1−r) + r^n = (1-r^{n+1})/(1-r).$

  • 3
    $\begingroup$ This is no different from the second method, except you skip a bunch of steps $\endgroup$
    – Justin
    Apr 23, 2013 at 18:48
  • $\begingroup$ yes... I miss the last paragraph... they are essentially the same... Alternatively, you can use induction to get the formula for $S_n$, suppose $1+r+\cdots+r^{n-1}= (1-r^n)/(1-r)$, for $n$, $1+r+\cdots+r^n= (1-r^n)/(1-r) + r^n = (1-r^{n+1})/(1-r)$, hope this helps. $\endgroup$
    – Ishigami
    Apr 23, 2013 at 18:53

By the Taylor development:

$$f(r)=\frac a{1-r}, f(0)=a$$

$$f'(r)=\frac a{(1-r)^2}, f'(0)=a$$

$$f''(r)=\frac{2a}{(1-r)^3}, f'(0)=2a$$

$$f'''(r)=\frac{6a}{(1-r)^4}, f'(0)=6a$$ $$\cdots$$ $$f^{(n)}(r)=\frac{n!a}{(1-r)^{n+1}}, f'(0)=n!a$$ $$\cdots$$


$$f(r)=a+ar+a\frac22 r^2+ a\frac{3!}{3!} r^3+\cdots a\frac{n!}{n!} r^3+\cdots$$


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