that must be taken so that the sum of these terms exceeds \(99.99\%\) of the sum to infinity of the series.

The sum of \(n\) terms of a GP is \(S_n = \dfrac{a(1-r^n)}{1-r}.\) For the GP above, \(a = 1\) and \(r = \dfrac{2}{3}\).

Since in our case \(\vert r \vert < 1,\) we can say the sum to infinity of the GP is \(S_\infty = \dfrac{1}{1-2/3} = 3.\)

We need to find the smallest \(n\) so that \(S_n > \dfrac{99.99 \times 3}{100}= 2.9997.\)

Let’s try to solve \(S_n = 2.9997.\) We have

\[S_n = \dfrac{1\times(1-(2/3)^n)}{1-2/3} = 3\left(1-\left(\dfrac{2}{3}\right)^n\right) = 2.9997.\]

Rearranging now tells us \(\left(\dfrac{2}{3}\right)^n = 0.0001.\) Taking logs to base \(e\), we have

\[n = \dfrac{\ln 0.0001}{\ln (2/3)} = 22.7....,\]

and so the value of \(n\) we seek is \(23\).

Alternatively, we could solve \(\left(\dfrac{2}{3}\right)^n = 0.0001\) by taking logs to base \(\dfrac{2}{3}\).

If you haven’t studied logs yet, you could find the threshold where \(\left(\dfrac{2}{3}\right)^n\) becomes small enough by trialling different \(n\) values.