# derivation of geometric series summation rule?

The sum of an infinite geometric series can be solved with the below equation, given that the common ratio, $$r$$, is bounded $$-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?

$$\Sigma = \frac{a_1}{1 - r}$$

• 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? Apr 16, 2021 at 1:37
• @AndrewD.Hwang, thx & apologies, I'm more often found on cross validated! Apr 16, 2021 at 1:50

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 $$0 we can tell that $$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 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;

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

• 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 Apr 16, 2021 at 1:42