Review question

# When does $1 + 2/3 + (2/3)^2 + \dotsb$ first exceed $0.9999S_\infty$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9185

## Solution

Find the least number of terms of the geometric series $\begin{equation*} 1 + \frac{2}{3} + \left( \frac{2}{3} \right)^2 + \dotsb \end{equation*}$

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.