Answer
We prove:
$G\left( k \right) = \dfrac{\pi }{2} - \dfrac{\pi }{2}\mathop \sum \limits_{n = 1}^\infty {\left( {\dfrac{{1\cdot3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot\cdot\cdot4\cdot\left( {2n} \right)}}} \right)^2}\dfrac{{{k^{2n}}}}{{2n - 1}}$, ${\ \ \ }$ converges for $\left| k \right| \lt 1$.
Work Step by Step
By Theorem 3:
${\left( {1 + x} \right)^a} = 1 + \dfrac{a}{{1!}}x + \dfrac{{a\left( {a - 1} \right)}}{{2!}}{x^2} + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot + \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n} + \cdot\cdot\cdot$
${\left( {1 + x} \right)^a} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$, ${\ \ \ }$ converges for $\left| x \right| \lt 1$.
So,
$\sqrt {1 + x} = {\left( {1 + x} \right)^{1/2}} = 1 + \dfrac{1}{2}x + \dfrac{{\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)}}{{2!}}{x^2} + \dfrac{{\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)}}{{3!}}{x^3} + \dfrac{{\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)}}{{4!}}{x^4} + \cdot\cdot\cdot$
$\sqrt {1 + x} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{1/2}\\
n
\end{array}} \right){x^n}$
$\sqrt {1 + x} = 1 + \dfrac{1}{2}x - \dfrac{1}{{2\cdot 4}}{x^2} + \dfrac{{1\cdot 3}}{{2\cdot 4\cdot 6}}{x^3} - \dfrac{{1\cdot 3\cdot 5}}{{2\cdot 4\cdot 6\cdot 8}}{x^4} + \cdot\cdot\cdot$
$\sqrt {1 + x} = 1 + \dfrac{1}{2}x - \dfrac{1}{3}\dfrac{{1\cdot 3}}{{2\cdot 4}}{x^2} + \dfrac{1}{5}\dfrac{{1\cdot 3\cdot 5}}{{2\cdot 4\cdot 6}}{x^3} - \dfrac{1}{7}\dfrac{{1\cdot 3\cdot 5\cdot 7}}{{2\cdot 4\cdot 6\cdot 8}}{x^4} + \cdot\cdot\cdot$
$\sqrt {1 + x} = 1 + \mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1}}\dfrac{{1\cdot 3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot 4\cdot 6\cdot\cdot\cdot\left( {2n} \right)\left( {2n - 1} \right)}}{x^n}$
Substituting $ - {k^2}{\sin ^2}t$ for $x$ in the series above gives
$\sqrt {1 - {k^2}{{\sin }^2}t} = 1 + \mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1}}\dfrac{{1\cdot3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot4\cdot6\cdot\cdot\cdot\left( {2n} \right)\left( {2n - 1} \right)}}{\left( { - {k^2}{{\sin }^2}t} \right)^n}$,
converges for $\left| { - {k^2}{{\sin }^2}t} \right| \lt 1$ or $\left| k \right| \lt 1$.
$\sqrt {1 - {k^2}{{\sin }^2}t} = 1 + \mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^{2n - 1}}\dfrac{{1\cdot 3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot 4\cdot 6\cdot\cdot\cdot\left( {2n} \right)\left( {2n - 1} \right)}}{k^{2n}}{\sin ^{2n}}t$
Take the definite integral:
$G\left( k \right) = \mathop \smallint \limits_0^{\pi /2} \sqrt {1 - {k^2}{{\sin }^2}t} {\rm{d}}t$
(1) ${\ \ \ \ \ }$ $G\left( k \right) = \dfrac{\pi }{2} + \mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^{2n - 1}}\dfrac{{1\cdot3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot4\cdot6\cdot\cdot\cdot\left( {2n} \right)\left( {2n - 1} \right)}}{k^{2n}}\mathop \smallint \limits_0^{\pi /2} {\sin ^{2n}}t{\rm{d}}t$
Using the result in Exercise 80 in Section 8.2, that is
$\mathop \smallint \limits_0^{\pi /2} {\sin ^{2n}}t{\rm{d}}t = \left( {\dfrac{{1\cdot 3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot 4\cdot 6\cdot\cdot\cdot\left( {2n} \right)\left( {2n - 1} \right)}}} \right)\dfrac{\pi }{2}$,
equation (1) becomes
$G\left( k \right) = \dfrac{\pi }{2} + \dfrac{\pi }{2}\mathop \sum \limits_\infty ^{n = 1} {\left( { - 1} \right)^{2n - 1}}{\left( {\dfrac{{1\cdot 3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot 4\cdot 6\cdot\cdot\cdot\left( {2n} \right)}}} \right)^2}\dfrac{{{k^{2n}}}}{{\left( {2n - 1} \right)}}$
Since ${\left( { - 1} \right)^{2n - 1}} = - 1$, therefore
$G\left( k \right) = \dfrac{\pi }{2} - \dfrac{\pi }{2}\mathop \sum \limits_{n = 1}^\infty {\left( {\dfrac{{1\cdot3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot\cdot\cdot4\cdot\left( {2n} \right)}}} \right)^2}\dfrac{{{k^{2n}}}}{{2n - 1}}$, ${\ \ \ }$ converges for $\left| k \right| \lt 1$.
