The sum of an infinite geometric series can be solved with the below equation, given that the common ratio, $r$, is bounded $ -1 <r< 1 $. I'm curious, is there a plain English explanation for why this works? If the explanation isn't quite so "plain English", how was this rule derived?
$\Sigma = \frac{a_1}{1 – r} $
Best Answer
A geometric series has a general nth term of
$$a_n=ar^{n-1}$$
Where $r$ is the common ratio, $a_n$ is the nth term and $a$ is the first term of the series
Hence the sum of a geometric series is
$$S_n=a+ar+ar^2....+ar^{n-1}$$
Multiplying the entire sum $S_n$ by the common ratio $r$
$$rS_n=ar+ar^2+ar^3....+ar^{n-1}+ar^n$$
For the case where $0<r<1$ we can tell that $S_n>rS_n$ , subtracting $rS_n$ from $S_n$ will lead to all terms cancelling out except for $a$ which is only present in $S_n$ and $ar^n$ which is only present in $rS_n$ hence
$$S_n=a+ar+ar^2....+ar^{n-1}$$ $$-(rS_n=ar+ar^2+ar^3....+ar^{n-1}+ar^n)$$
$$\implies S_n-rS_n= a-ar^n$$
Simplifying it further by taking out the common factors
$$S_n(1-r)=a(1-r^n) \therefore S_n=\frac{a(1-r^n)}{1-r}$$
Now that we have gotten the formula of the sum of a geometric series we can derive that for the sum of an infinite geometric series by noting that if some number $k$ is such that $-1<k<1$ then
$$\lim_{n \to \infty} k^n=0 $$
If you want to prove this to yourself try multiplying $0.5$ by itself and note how the resulting answer is smaller than $0.5$
Now we can use the earlier result and our formula if $-1<r<1$;
$$\lim_{n \to \infty} S_n=\frac{a(1-r^n)}{1-r} \therefore S_{\infty}=\frac{a}{1-r}$$