Answer
(a) The series converges if $ - 1 \lt x \lt 1$; $S = \dfrac{x}{{1 + {x^2}}}$.
(b) The series converges if $\left| x \right| \gt 2$; $S = \dfrac{1}{{{x^2} - 2x}}$.
(c) The series converges if $x \gt 0$; $S = \dfrac{1}{{{{\rm{e}}^x} - 1}}$.
Work Step by Step
(a) Write
$\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k + 1}} = x - {x^3} + {x^5} - {x^7} + {x^9} - \cdot\cdot\cdot$
$\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k + 1}} = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( {{x^2}} \right)^k}x = x\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( {{x^2}} \right)^k}$
According to Exercise 30 (c):
$\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( {{x^2}} \right)^k} = \dfrac{1}{{1 + {x^2}}}$ ${\ \ \ }$ if $ - 1 \lt x \lt 1$
Therefore
$\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k + 1}} = x\left( {\dfrac{1}{{1 + {x^2}}}} \right)$ ${\ \ \ }$ if $ - 1 \lt x \lt 1$
Thus, the series converges if $ - 1 \lt x \lt 1$ and its sum is $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k + 1}} = \dfrac{x}{{1 + {x^2}}}$.
(b) Write
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \dfrac{1}{{{x^2}}} + \dfrac{2}{{{x^3}}} + \dfrac{4}{{{x^4}}} + \dfrac{8}{{{x^5}}} + \dfrac{{16}}{{{x^6}}} + \cdot\cdot\cdot$
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{x^2}}}{\left( {\dfrac{2}{x}} \right)^k}$
The series on the right-hand side converges when it is a geometric series with $a = \dfrac{1}{{{x^2}}}$ and $\left| r \right| = \left| {\dfrac{2}{x}} \right| \lt 1$. That is,
$\left| x \right| \gt 2$
And its sum by 9.3.3 Theorem is
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{x^2}}}{\left( {\dfrac{2}{x}} \right)^k} = \dfrac{{\dfrac{1}{{{x^2}}}}}{{1 - \dfrac{2}{x}}} = \dfrac{1}{{{x^2} - 2x}}$
Therefore,
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \dfrac{1}{{{x^2} - 2x}}$ ${\ \ \ }$ if $\left| x \right| \gt 2$
Thus, the series converges if $\left| x \right| \gt 2$ and its sum is $\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \dfrac{1}{{{x^2} - 2x}}$.
(c) Write
$\mathop \sum \limits_{k = 0}^\infty {{\rm{e}}^{ - \left( {k + 1} \right)x}} = {{\rm{e}}^{ - x}} + {{\rm{e}}^{ - 2x}} + {{\rm{e}}^{ - 3x}} + {{\rm{e}}^{ - 4x}} + {{\rm{e}}^{ - 5x}} + \cdot\cdot\cdot$
$\mathop \sum \limits_{k = 0}^\infty {{\rm{e}}^{ - \left( {k + 1} \right)x}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{{\rm{e}}^{\left( {k + 1} \right)x}}}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{{\rm{e}}^x}}}{\left( {\dfrac{1}{{{{\rm{e}}^x}}}} \right)^k}$
The series on the right-hand side converges when it is a geometric series with $a = \dfrac{1}{{{{\rm{e}}^x}}}$ and $\left| r \right| = \left| {\dfrac{1}{{{{\rm{e}}^x}}}} \right| \lt 1$. That is,
${{\rm{e}}^x} \gt 1$
$\ln {{\rm{e}}^x} \gt \ln 1 = 0$
$x \gt 0$
And the sum by 9.3.3 Theorem is
$\mathop \sum \limits_{k = 0}^\infty {{\rm{e}}^{ - \left( {k + 1} \right)x}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{{\rm{e}}^x}}}{\left( {\dfrac{1}{{{{\rm{e}}^x}}}} \right)^k} = \dfrac{{\dfrac{1}{{{{\rm{e}}^x}}}}}{{1 - \dfrac{1}{{{{\rm{e}}^x}}}}} = \dfrac{1}{{{{\rm{e}}^x} - 1}}$
Thus, the series converges if $x \gt 0$ and its sum is $\mathop \sum \limits_{k = 0}^\infty {{\rm{e}}^{ - \left( {k + 1} \right)x}} = \dfrac{1}{{{{\rm{e}}^x} - 1}}$.