Answer
(a) $\dfrac{4}{3}$
(b) $ - \dfrac{3}{4}$
Work Step by Step
(a) By Theorem 9.4.3, the original series can be rewritten as
$\left( {\dfrac{1}{2} + \dfrac{1}{4}} \right) + \left( {\dfrac{1}{{{2^2}}} + \dfrac{1}{{{4^2}}}} \right) + \cdot\cdot\cdot + \left( {\dfrac{1}{{{2^k}}} + \dfrac{1}{{{4^k}}}} \right) + \cdot\cdot\cdot$
$ = \left( {\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \cdot\cdot\cdot + \dfrac{1}{{{2^k}}} + \cdot\cdot\cdot} \right) + \left( {\dfrac{1}{4} + \dfrac{1}{{{4^2}}} + \cdot\cdot\cdot + \dfrac{1}{{{4^k}}} + \cdot\cdot\cdot} \right)$
$ = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{2}{\left( {\dfrac{1}{2}} \right)^k} + \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{4}{\left( {\dfrac{1}{4}} \right)^k}$
Both the series on the right-hand side are geometric. Thus, by Theorem 9.3.3, we obtain
$\left( {\dfrac{1}{2} + \dfrac{1}{4}} \right) + \left( {\dfrac{1}{{{2^2}}} + \dfrac{1}{{{4^2}}}} \right) + \cdot\cdot\cdot + \left( {\dfrac{1}{{{2^k}}} + \dfrac{1}{{{4^k}}}} \right) + \cdot\cdot\cdot$
$ = \dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{2}}} + \dfrac{{\dfrac{1}{4}}}{{1 - \dfrac{1}{4}}} = 1 + \dfrac{1}{3} = \dfrac{4}{3}$
Thus,
$\left( {\dfrac{1}{2} + \dfrac{1}{4}} \right) + \left( {\dfrac{1}{{{2^2}}} + \dfrac{1}{{{4^2}}}} \right) + \cdot\cdot\cdot + \left( {\dfrac{1}{{{2^k}}} + \dfrac{1}{{{4^k}}}} \right) + \cdot\cdot\cdot = \dfrac{4}{3}$
(b) By Theorem 9.4.3:
$\mathop \sum \limits_{k = 1}^\infty \left( {\dfrac{1}{{{5^k}}} - \dfrac{1}{{k\left( {k + 1} \right)}}} \right) = \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{5^k}}} - \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k\left( {k + 1} \right)}}$
1. By Theorem 9.3.3, the geometric series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{5^k}}}$ has the sum
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{5^k}}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{5}{\left( {\dfrac{1}{5}} \right)^k} = \dfrac{{\dfrac{1}{5}}}{{1 - \dfrac{1}{5}}} = \dfrac{1}{4}$
2. According to Example 5 in Section 9.3:
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k\left( {k + 1} \right)}} = 1$
Therefore,
$\mathop \sum \limits_{k = 1}^\infty \left( {\dfrac{1}{{{5^k}}} - \dfrac{1}{{k\left( {k + 1} \right)}}} \right) = \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{5^k}}} - \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k\left( {k + 1} \right)}} = \dfrac{1}{4} - 1 = - \dfrac{3}{4}$
