Answer
The upper bound on the absolute error is ${a_{11}} \approx 0.01734$.
${s_{10}}$ as compared to the exact sum $S$:
$\left| {S - {s_{10}}} \right| = 0.0104 \lt 0.01734$
Work Step by Step
The series can be written as $\mathop \sum \limits_{k = 1}^\infty {\left( { - 1} \right)^{k - 1}}{\left( {\dfrac{2}{3}} \right)^{k - 1}}$, where ${a_k} = {\left( {\dfrac{2}{3}} \right)^{k - 1}}$.
1. According to Theorem 9.6.2, the upper bound on the absolute error is ${a_{n + 1}}$, given by Eq. (4):
$\left| {S - {s_n}} \right| \le {a_{n + 1}}$
For $n=10$, we have
$\left| {S - {s_{10}}} \right| \le {a_{11}}$
Thus, the upper bound on the absolute error is ${a_{11}} = {\left( {\dfrac{2}{3}} \right)^{10}} = \dfrac{{1024}}{{59049}} \approx 0.01734$.
2. Compute ${s_{10}}$ and the exact sum $S$:
${s_{10}} = 1 - \dfrac{2}{3} + \dfrac{4}{9} - \dfrac{8}{{27}} + \dfrac{{16}}{{81}} - \dfrac{{32}}{{243}} + \dfrac{{64}}{{729}} - \dfrac{{128}}{{2187}} + \dfrac{{256}}{{6561}} - \dfrac{{512}}{{19683}} = \dfrac{{11605}}{{19683}} \approx 0.5896$
Write
$S = \mathop \sum \limits_{k = 1}^\infty {\left( { - 1} \right)^{k - 1}}{\left( {\dfrac{2}{3}} \right)^{k - 1}} = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( {\dfrac{2}{3}} \right)^k} = \mathop \sum \limits_{k = 0}^\infty {\left( { - \dfrac{2}{3}} \right)^k}$
According to Theorem 9.3.3, the exact sum $S$ is
$S = \dfrac{1}{{1 - \left( { - \dfrac{2}{3}} \right)}} = \dfrac{3}{5} = 0.6$
As a verification of the accuracy, we compare ${s_{10}}$ and S:
$\left| {S - {s_{10}}} \right| \le {a_{11}}$
$\left| {S - {s_{10}}} \right| = \left| {0.6 - 0.5896} \right| = 0.0104 \lt 0.01734$
