Chapter 9 - Infinite Series - 9.6 Alternating Series; Absolute And Conditional Convergence - Exercises Set 9.6 - Page 647: 37

$n=9999$

We have ${a_k} = \dfrac{1}{k}$. 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 $0.0001$, we must have ${a_{n + 1}} \le 0.0001$ $\dfrac{1}{{n + 1}} \le \dfrac{1}{{10000}}$ $n + 1 \ge 10000$ $n \ge 9999$ We can take $n=9999$.