# Formula for an infinite geometric series (Part 10)

I conclude this series of posts by considering the formula for an infinite geometric series. Somewhat surprisingly (to students), the formula for an infinite geometric series is actually easier to remember than the formula for a finite geometric series.

One way of deriving the formula parallels the derivation for a finite geometric series. If $a_1, a_2, a_3, \dots$ are the first terms of an infinite geometric sequence, let $S = a_1 + a_2 + a_3 + \dots$

Recalling the formula for an geometric sequence, we know that $a_2 = a_1 r$ $a_3 = a_1 r^2$ $\vdots$

Substituting, we find $S = a_1 + a_1 r+ a_1 r^2 \dots$

Once again, we multiply both sides by $-r$. $-rS = -a_1r - a_1 r^2- a_1 r^3 \dots$

Next, we add the two equations. Notice that almost everything cancels on the right-hand side… except for the leading term $a_1$.  (Unlike yesterday’s post, there is no “last” term that remains since the series is infinite.) Therefore, $S - rS = a_1$ $S(1-r) = a_1$ $S = \displaystyle \frac{a_1}{1-r}$

A quick pedagogical note: I find that this derivation “sells” best to students when I multiply by $-r$ and add, as opposed to multiplying by $r$ and subtracting. The above derivation is helpful for remembering the formula but glosses over an extremely important detail: not every infinite geometric series converges. For example, if $a_1 = 1$ and $r = 2$, then the infinite geometric series becomes $1 + 2 + 4 + 8 + 16 + \dots$,

which clearly does not have a finite answer. We say that this series diverges. In other words, trying to evaluate this sum makes as much sense as trying to divide a number by zero: there is no answer.

That said, it can be shown that, as long as $-1 < r < 1$, then the above geometric series converges, so that $a_1 + a_1 r + a_1 r^2 + \dots = \displaystyle \frac{a_1}{1-r}$

The formal proof requires the use of the formula for a finite geometric series: $a_1 + a_1 r + a_1 r^2 + \dots + a_1 r^{n-1} = \displaystyle \frac{a_1(1-r^n)}{1-r}$

We then take the limit as $n \to \infty$: $\displaystyle \lim_{n \to \infty} a_1 + a_1 r + a_1 r^2 + \dots + a_1 r^{n-1} = \displaystyle \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r}$ $a_1 + a_1 r + a_1 r^2 + \dots = \displaystyle \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r}$

On the right-hand side, the only piece that contains an $n$ is the term $r^n$. If $-1 < r < 1$, then $r^n \to 0$ as $n \to \infty$. (This limit fails, however, if $r \ge 1$ or $r \le -1$.) Therefore, $a_1 + a_1 r + a_1 r^2 + \dots = \displaystyle \lim_{n \to \infty} \frac{a_1(1-0)}{1-r} = \displaystyle \frac{a_1}{1-r}$

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