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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} $

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    $\begingroup$ Formally, write $S = a + ar + ar^{2} + ar^{3} + \cdots$. Multiply both sides by $r$ (noting that the terms "shift by one") and subtract. Every term except the constant $a$ cancels, so $(1 - r)S = a$, or $S = a/(1-r)$. This is surely answered somewhere on site; have you searched, and did you not find anything? $\endgroup$ Apr 16, 2021 at 1:37
  • $\begingroup$ @AndrewD.Hwang, thx & apologies, I'm more often found on cross validated! $\endgroup$
    – jbuddy_13
    Apr 16, 2021 at 1:50

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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 $-1<r<1$;

$$\lim_{n \to \infty} S_n=\frac{a(1-r^n)}{1-r} \therefore S_{\infty}=\frac{a}{1-r}$$

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    $\begingroup$ Small typo: you wrote $1+r^n$ instead of $1-r^n$ while solving for $S_n$. Other than that, this hits the nail on the head +1 $\endgroup$ Apr 16, 2021 at 1:42

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