Answer
(a) Converges; $S=\frac{{18}}{5}$
(b) Diverges
(c) Converges; $S= \frac{1}{2}$
Work Step by Step
(a) $\mathop \sum \limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1}}{2^k}\cdot{3^{2 - k}}$
1. By CAS:
$\mathop \sum \limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1}}{2^k}\cdot{3^{2 - k}} = \dfrac{{18}}{5}$
2. By hand calculation:
Write
$\mathop \sum \limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1}}{2^k}\cdot{3^{2 - k}} = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^{k + 2}}{2^{k + 1}}\cdot{3^{1 - k}}$
$ = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}\left( 2 \right)\left( 3 \right){\left( {\dfrac{2}{3}} \right)^k}$
So, $\mathop \sum \limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1}}{2^k}\cdot{3^{2 - k}} = \mathop \sum \limits_{k = 0}^\infty 6{\left( { - \dfrac{2}{3}} \right)^k}$.
This is a geometric series with $a=3$ and $\left| r \right| = \left| { - \dfrac{2}{3}} \right| \lt 1$.
The sum by 9.3.3 Theorem is
$\mathop \sum \limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1}}{2^k}\cdot{3^{2 - k}} = \dfrac{6}{{1 - \left( { - \dfrac{2}{3}} \right)}} = \dfrac{{18}}{5}$
This confirms the result. The series converges.
(b) $\mathop \sum \limits_{k = 1}^\infty \dfrac{{{3^{3k}}}}{{{5^{k - 1}}}}$
1. By CAS:
The series: $\mathop \sum \limits_{k = 1}^\infty \dfrac{{{3^{3k}}}}{{{5^{k - 1}}}}$ diverges.
2. By hand calculation:
Write
$\mathop \sum \limits_{k = 1}^\infty \dfrac{{{3^{3k}}}}{{{5^{k - 1}}}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{{{3^{3k + 3}}}}{{{5^k}}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{{27{{\left( {{3^3}} \right)}^k}}}{{{5^k}}} = \mathop \sum \limits_{k = 0}^\infty 27{\left( {\dfrac{{27}}{5}} \right)^k}$
This is a geometric series with $a=27$ and $\left| r \right| = \dfrac{{27}}{5} \gt 1$. By 9.3.3 Theorem, the series diverges.
This confirms the result.
(c) $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{4{k^2} - 1}}$
1. By CAS:
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{4{k^2} - 1}} = \dfrac{1}{2}$
2. By hand calculation:
Write
$\dfrac{1}{{4{k^2} - 1}} = \dfrac{1}{2}\left( {\dfrac{1}{{2k - 1}} - \dfrac{1}{{2k + 1}}} \right)$
So, we have
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{4{k^2} - 1}} = \dfrac{1}{2}\mathop \sum \limits_{k = 1}^\infty \left( {\dfrac{1}{{2k - 1}} - \dfrac{1}{{2k + 1}}} \right)$
The $n$th partial sum is
$\mathop \sum \limits_{k = 1}^n \dfrac{1}{{4{k^2} - 1}} = \dfrac{1}{2}\left[ {\left( {1 - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{5}} \right) + \left( {\dfrac{1}{5} - \dfrac{1}{7}} \right) + \cdot\cdot\cdot + \left( {\dfrac{1}{{2n - 1}} - \dfrac{1}{{2n + 1}}} \right)} \right]$
$\mathop \sum \limits_{k = 1}^n \dfrac{1}{{4{k^2} - 1}} = \dfrac{1}{2}[(1 + \left( { - \dfrac{1}{3} + \dfrac{1}{3}} \right) + \left( { - \dfrac{1}{5} + \dfrac{1}{5}} \right) + \cdot\cdot\cdot + \left( { - \dfrac{1}{{2n - 1}} + \dfrac{1}{{2n - 1}}} \right) - \dfrac{1}{{2n + 1}}]$
$\mathop \sum \limits_{k = 1}^n \dfrac{1}{{4{k^2} - 1}} = \dfrac{1}{2}\left( {1 - \dfrac{1}{{2n + 1}}} \right)$
Taking the limit:
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{4{k^2} - 1}} = \dfrac{1}{2}\mathop {\lim }\limits_{n \to \infty } \left( {1 - \dfrac{1}{{2n + 1}}} \right) = \dfrac{1}{2}\left( {\mathop {\lim }\limits_{n \to \infty } 1 - \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{2n + 1}}} \right) = \dfrac{1}{2}$
This confirms the result. The series converges.
