Chapter 9 - Infinite Series - 9.6 Alternating Series; Absolute And Conditional Convergence - Exercises Set 9.6 - Page 647: 35

${a_{100}} = 0.1$

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 $n=99$, the upper bound on the absolute error is ${a_{100}} = \dfrac{1}{{\sqrt {100} }} = \dfrac{1}{{10}} = 0.1$.