Infinite Geometric Series Formula Derivation

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

Solution 1:

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.

Solution 2:

$a+ar+a(r^{2})...=s$

$a+rs=s$

$a+rs-s=0$

$a+(r-1)s=0$

$-a=(r-1)s$

$\frac{-a}{r-1}=s$

$r-1=-(1-r)$

$\frac{-a}{-(1-r)}=s$

$s=\frac{a}{1-r}$

Solution 3:

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.

Solution 4:

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 $n$ , $1+r+\cdots+r^{n−1}+r^n =(1−r^n )/(1−r) + r^n = (1-r^{n+1})/(1-r).$