Answer
See proof
Work Step by Step
(a) We have the series $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^k}$.
Write
$\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^k} = \mathop \sum \limits_{k = 0}^\infty {\left( { - x} \right)^k}$
If $ - 1 \lt \left( { - x} \right) \lt 1$, by 9.3.3 Theorem the series $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^k} = \mathop \sum \limits_{k = 0}^\infty {\left( { - x} \right)^k}$ converges, where $a=1$ and $\left| r \right| = \left| { - x} \right| \lt 1$. Thus,
$\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^k} = \mathop \sum \limits_{k = 0}^\infty {\left( { - x} \right)^k} = \dfrac{1}{{1 - \left( { - x} \right)}} = \dfrac{1}{{1 + x}}$
Hence, $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^k} = \dfrac{1}{{1 + x}}$ ${\ \ \ }$ if $ - 1 \lt x \lt 1$.
(b) We have the series $\mathop \sum \limits_{k = 0}^\infty {\left( {x - 3} \right)^k}$.
Write $r=x-3$. So, the series can be rewritten as
$\mathop \sum \limits_{k = 0}^\infty {\left( {x - 3} \right)^k} = \mathop \sum \limits_{k = 0}^\infty {r^k}$
If $ - 1 \lt r \lt 1$, by 9.3.3 Theorem the series $\mathop \sum \limits_{k = 0}^\infty {\left( {x - 3} \right)^k} = \mathop \sum \limits_{k = 0}^\infty {r^k}$ converges, where $a=1$ and $\left| r \right| = \left| {x - 3} \right| \lt 1$. Thus,
$\mathop \sum \limits_{k = 0}^\infty {\left( {x - 3} \right)^k} = \mathop \sum \limits_{k = 0}^\infty {r^k} = \dfrac{1}{{1 - \left( {x - 3} \right)}} = \dfrac{1}{{4 - x}}$
Since $\left| r \right| = \left| {x - 3} \right| \lt 1$, so
$ - 1 \lt x - 3 \lt 1$
$2 \lt x \lt 4$
Hence, $\mathop \sum \limits_{k = 0}^\infty {\left( {x - 3} \right)^k} = \dfrac{1}{{4 - x}}$ ${\ \ \ }$ if $2 \lt x \lt 4$.
(c) We have the series $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k}}$.
Write
$\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k}} = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( {{x^2}} \right)^k} = \mathop \sum \limits_{k = 0}^\infty {\left( { - {x^2}} \right)^k}$
If $ - 1 \lt \left( { - {x^2}} \right) \lt 1$, by 9.3.3 Theorem the series $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k}} = \mathop \sum \limits_{k = 0}^\infty {\left( { - {x^2}} \right)^k}$ converges, where $a=1$ and $\left| r \right| = \left| { - {x^2}} \right| \lt 1$. Thus,
$\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k}} = \mathop \sum \limits_{k = 0}^\infty {\left( { - {x^2}} \right)^k} = \dfrac{1}{{1 - \left( { - {x^2}} \right)}} = \dfrac{1}{{1 + {x^2}}}$
Since $\left| r \right| = \left| { - {x^2}} \right| \lt 1$, so $ - 1 \lt x \lt 1$.
Hence, $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k}} = \dfrac{1}{{1 + {x^2}}}$ ${\ \ \ }$ if $ - 1 \lt x \lt 1$.
