A calculus of the absurd

3.4 Geometric series

A geometric series is anything which looks like

\begin{equation} \label {geometric series} a + ar + ar^2 + ... + ar^{n-1} \end{equation}

Here, \(a\) is the starting term, \(r\) is called the "common ration" (this name might make a bit more sense if you consider how each term is \(r\) times smaller than the next term) and \(n\) is the total number of terms. We can write the series in \(\sum \)-notation as (it’s worth expanding this and checking that it is indeed the same as Equation 3.26)

\begin{equation} \sum _{j=1}^{n} \left [ ar^{j-1} \right ] \end{equation}

3.4.1 Finding the sum of a geometric series

This is one of those things which is much harder to find than it is to verify that one has indeed found the correct thing. For A Level maths the sum is given in the formula booklet, and it’s not necessary to be able to show why it is true.

If we call the value of Equation 3.26 \(S_n\) (read as "the sum of \(n\) terms of the series), then we can multiply \(S_n\) by \(r\) and get that

\begin{equation} r S_n = ar + ar^2 + ... + ar^{n} \end{equation}

From here we can proceed by finding the value of \(S_n - r S_n\)

\begin{align} S_n - r S_n &= (a + ar + ar^2 + ... + ar^{n-1}) - (ar + ar^2 + ... + ar^{n}) \\ &= a - ar^{n} \end{align}

This can then be rearranged a bit by factoring out the \(S_n\) and then dividing through by \(1-r\).

\begin{align} & S_n(1 - n) = a - ar^{n} \\ & S_n = \frac {a - ar^{n}}{1-r} \\ & S_n = \frac {a(1 - r^{n})}{1-r} \label {sum of a geometric series} \end{align}