Answer
${s_5} \approx 0.4073$
Work Step by Step
Write
$S = \dfrac{1}{{1\cdot2}} - \dfrac{1}{{2\cdot{2^2}}} + \dfrac{1}{{3\cdot{2^3}}} - \dfrac{1}{{4\cdot{2^4}}} + \cdot\cdot\cdot = \mathop \sum \limits_{k = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{k - 1}}}}{{k\cdot{2^k}}}$,
where ${a_k} = \dfrac{1}{{k\cdot{2^k}}}$.
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 two decimal-place accuracy, we must choose a value of $n$ such that
$\left| {S - {s_n}} \right| \le {a_{n + 1}} \lt 0.005$
$\dfrac{1}{{\left( {n + 1} \right)\cdot{2^{n + 1}}}} \lt 0.005$
$\left( {n + 1} \right){2^{n + 1}} \gt 200$
Using a calculating utility, we obtain $n \gt 4.25$. Thus, we can take $n=5$.
Thus, the approximated sum is ${s_5}$:
${s_5} = \dfrac{1}{{1\cdot2}} - \dfrac{1}{{2\cdot{2^2}}} + \dfrac{1}{{3\cdot{2^3}}} - \dfrac{1}{{4\cdot{2^4}}} + \dfrac{1}{{5\cdot{2^5}}} = \dfrac{{391}}{{960}} \approx 0.4073$
