Answer
The upper bound on the absolute error is ${a_{11}} \approx 0.00073$.
${s_{10}}$ as compared to the exact sum $S$:
$\left| {S - {s_{10}}} \right| = 0.0005 \lt 0.00073$
Work Step by Step
The series can be written as $\mathop \sum \limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1}}\dfrac{3}{{{2^{k + 1}}}}$, where ${a_k} = \dfrac{3}{{{2^{k + 1}}}}$.
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 ${a_{11}} = \dfrac{3}{{{2^{12}}}} = \dfrac{3}{{4096}} \approx 0.00073$.
2. Compute ${s_{10}}$ and the exact sum $S$:
${s_{10}} = \dfrac{3}{4} - \dfrac{3}{8} + \dfrac{3}{{16}} - \dfrac{3}{{32}} + \dfrac{3}{{64}} - \dfrac{3}{{128}} + \dfrac{3}{{256}} - \dfrac{3}{{512}} + \dfrac{3}{{1024}} - \dfrac{3}{{2048}} = \dfrac{{1023}}{{2048}} \approx 0.4995$
Write
$S = \mathop \sum \limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1}}\dfrac{3}{{{2^{k + 1}}}} = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^{k + 2}}\dfrac{3}{{{2^{k + 2}}}} = \dfrac{3}{4}\mathop \sum \limits_{k = 0}^\infty {\left( { - \dfrac{1}{2}} \right)^k}$
According to Theorem 9.3.3, the exact sum $S$ is
$S = \dfrac{3}{4}\left( {\dfrac{1}{{1 - \left( { - \dfrac{1}{2}} \right)}}} \right) = \dfrac{3}{4}\left( {\dfrac{2}{3}} \right) = \dfrac{1}{2} = 0.5$
As a verification of the accuracy, we compare ${s_{10}}$ and $S$:
$\left| {S - {s_{10}}} \right| \le {a_{11}}$
$\left| {S - {s_{10}}} \right| = \left| {0.5 - 0.4995} \right| = 0.0005 \lt 0.00073$
