Answer
4 terms are needed.
Using a calculator, we compute:
${S_4} \approx 0.182267$, ${\ \ \ \ \ }$ $\ln 1.2 \approx 0.182322$
Hence, $\left| {\ln 1.2 - {S_4}} \right| \approx 0.0000549 \lt 0.0001$
Work Step by Step
From Table 2 we get
$S = \ln \left( {1 + x} \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{x^n}}}{n} = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \lt 1$ and $x=1$.
We have for this alternating series, the positive terms: ${b_n} = \dfrac{{{x^n}}}{n}$.
Let $S = \ln \left( {1 + x} \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{x^n}}}{n}$.
By Eq. (2) in Section 11.4:
$\left| {S - {S_n}} \right| \lt {b_{n + 1}}$
We choose $n$ so that
$\left| {S - {S_n}} \right| \lt {b_{n + 1}} \lt {10^{ - 4}}$
For $x = 0.2$, we have
$\dfrac{{{{\left( {\dfrac{1}{5}} \right)}^{n + 1}}}}{{n + 1}} \lt {10^{ - 4}}$
${5^{n + 1}}\left( {n + 1} \right) > {10^4}$
We evaluate and list some values of $n$ in the following table:
$\begin{array}{*{20}{c}}
n&{}&{{5^{n + 1}}\left( {n + 1} \right)}\\
2&{}&{375}\\
3&{}&{2500}\\
4&{}&{15625}\\
5&{}&{93750}
\end{array}$
Based on the results in the table above, we choose $N=4$ such that $\ln 1.2$ is within an error of at most $0.0001$.
Using a calculator, we compute:
${S_4} \approx 0.182267$, ${\ \ \ \ \ }$ $\ln 1.2 \approx 0.182322$
Hence, $\left| {\ln 1.2 - {S_4}} \right| \approx 0.0000549 \lt 0.0001$
