How do I work this out?
The first three terms of a geometric series are $2$, $-\frac{1}{2}$, $\frac{1}{8}$. Find the least value of $n$ for which the sum of the first $n$ terms of this series differs from the sum to infinity by less than $10^{-5}$.
So, it will be:
$$ S_{\infty}-S_n > 0.00001 \\ 1.6 - \frac{2\left(1 - \left(-\frac{1}{4}\right)^n\right)}{1.25} > 0.00001 $$
But I get stuck at this point. How do I commence?
On
Since it's an alternating series, you just need to find the first term where $|2(-\frac14)^n| < 10^{-5}$.
$$\begin{align} \left|2\left(-\frac14\right)^n)\right| &< 10^{-5}\\ 2\left(\frac14\right)^n &< 10^{-5}\\ \left(\frac14\right)^n &< \frac{1}{200000}\\ 200000 & < 4^n\\ \end{align}$$
$4^8 = 16384$ and $4^9 = 262144$, so the sum of the terms through $2\left(-\frac14\right)^8$ is accurate to within $10^{-5}$.
With help from André:
$$ 1.6 - 1.6\left(1 - \left(-\frac{1}{4}^n\right)\right) < 0.00001 \\ 1.6\left(\frac{1}{4}^n\right) < 0.00001 \\ \frac{1}{4}^n < 0.00000625 \\ n > \frac{\log 0.00000625}{\log 0.25} \\ n > 9 $$