Answer
$n=39999$
Work Step by Step
We have ${a_k} = \dfrac{1}{{\sqrt k }}$.
According to Theorem 9.6.2, the absolute error $\left| {S - {s_n}} \right|$ is given by Eq. (4):
$\left| {S - {s_n}} \right| \le {a_{n + 1}}$
For an accuracy to two decimal places, we have
${a_{n + 1}} \le 0.005$
Thus,
$\dfrac{1}{{\sqrt {n + 1} }} \le 0.005$
$\sqrt {n + 1} \ge 200$
$n + 1 \ge 40000$
$n \ge 39999$
Thus, we can take $n=39999$.
