Answer
(a) $\mathop \sum \limits_{k = 2}^\infty \left[ {\dfrac{1}{{{k^2} - 1}} - \dfrac{7}{{{{10}^{k - 1}}}}} \right] = - \dfrac{1}{{36}}$
(b) $\mathop \sum \limits_{k = 1}^\infty \left[ {{7^{ - k}}\cdot{3^{k + 1}} - \dfrac{{{2^{k + 1}}}}{{{5^k}}}} \right] = \dfrac{{11}}{{12}}$
Work Step by Step
(a)
(i) Consider the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{{k^2} - 1}}$.
We can write $\dfrac{1}{{{k^2} - 1}} = \dfrac{1}{2}\left( {\dfrac{1}{{k - 1}} - \dfrac{1}{{k + 1}}} \right)$. Thus, we have the $n$th partial sums:
${s_n} = \mathop \sum \limits_{k = 2}^n \dfrac{1}{{{k^2} - 1}} = \dfrac{1}{2}\mathop \sum \limits_{k = 2}^n \left( {\dfrac{1}{{k - 1}} - \dfrac{1}{{k + 1}}} \right)$
${s_n} = \dfrac{1}{2}\left[ {\left( {1 - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{2} - \dfrac{1}{4}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{5}} \right) + \cdot\cdot\cdot + \left( {\dfrac{1}{{n - 1}} - \dfrac{1}{{n + 1}}} \right)} \right]$
${s_n} = \dfrac{1}{2}\left[ {1 + \dfrac{1}{2} + \left( { - \dfrac{1}{3} + \dfrac{1}{3}} \right) + \left( { - \dfrac{1}{4} + \dfrac{1}{4}} \right) + \cdot\cdot\cdot + \left( { - \dfrac{1}{{n - 1}} + \dfrac{1}{{n - 1}}} \right) - \dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right]$
${s_n} = \mathop \sum \limits_{k = 2}^n \dfrac{1}{{{k^2} - 1}} = \dfrac{1}{2}\left( {\dfrac{3}{2} - \dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right)$
The sum of the series is
$\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{{k^2} - 1}} = \mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{3}{4} - \dfrac{1}{{2n}} - \dfrac{1}{{2\left( {n + 1} \right)}}} \right)$
$\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{{k^2} - 1}} = \dfrac{3}{4}$
(ii) Consider the geometric series $\mathop \sum \limits_{k = 2}^\infty \dfrac{7}{{{{10}^{k - 1}}}}$.
Write
$\mathop \sum \limits_{k = 2}^\infty \dfrac{7}{{{{10}^{k - 1}}}} = \mathop \sum \limits_{k = 2}^\infty 70{\left( {\dfrac{1}{{10}}} \right)^k} = \mathop \sum \limits_{k = 0}^\infty 70{\left( {\dfrac{1}{{10}}} \right)^{k + 2}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{7}{{10}}{\left( {\dfrac{1}{{10}}} \right)^k}$
By Theorem 9.3.3, the geometric series has the sum
$\mathop \sum \limits_{k = 2}^\infty \dfrac{7}{{{{10}^{k - 1}}}} = \dfrac{{\dfrac{7}{{10}}}}{{1 - \dfrac{1}{{10}}}} = \dfrac{7}{9}$
Therefore, by Theorem 9.4.3:
$\mathop \sum \limits_{k = 2}^\infty \left[ {\dfrac{1}{{{k^2} - 1}} - \dfrac{7}{{{{10}^{k - 1}}}}} \right] = \mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{{k^2} - 1}} - \mathop \sum \limits_{k = 2}^\infty \dfrac{7}{{{{10}^{k - 1}}}} = \dfrac{3}{4} - \dfrac{7}{9} = - \dfrac{1}{{36}}$
(b) By Theorem 9.4.3, write
$\mathop \sum \limits_{k = 1}^\infty \left[ {{7^{ - k}}\cdot{3^{k + 1}} - \dfrac{{{2^{k + 1}}}}{{{5^k}}}} \right] = \mathop \sum \limits_{k = 1}^\infty {7^{ - k}}\cdot{3^{k + 1}} - \mathop \sum \limits_{k = 1}^\infty \dfrac{{{2^{k + 1}}}}{{{5^k}}}$
$ = \mathop \sum \limits_{k = 1}^\infty 3{\left( {\dfrac{3}{7}} \right)^k} - \mathop \sum \limits_{k = 1}^\infty 2{\left( {\dfrac{2}{5}} \right)^k}$
$ = \mathop \sum \limits_{k = 0}^\infty 3{\left( {\dfrac{3}{7}} \right)^{k + 1}} - \mathop \sum \limits_{k = 0}^\infty 2{\left( {\dfrac{2}{5}} \right)^{k + 1}}$
$ = \mathop \sum \limits_{k = 0}^\infty \dfrac{9}{7}{\left( {\dfrac{3}{7}} \right)^k} - \mathop \sum \limits_{k = 0}^\infty \dfrac{4}{5}{\left( {\dfrac{2}{5}} \right)^k}$
By Theorem 9.3.3:
$\mathop \sum \limits_{k = 1}^\infty \left[ {{7^{ - k}}\cdot{3^{k + 1}} - \dfrac{{{2^{k + 1}}}}{{{5^k}}}} \right] = \dfrac{{\dfrac{9}{7}}}{{1 - \dfrac{3}{7}}} - \dfrac{{\dfrac{4}{5}}}{{1 - \dfrac{2}{5}}} = \dfrac{9}{4} - \dfrac{4}{3} = \dfrac{{11}}{{12}}$
