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.
