Answer
$S = \mathop \smallint \limits_0^1 \dfrac{{{\rm{d}}x}}{{\sqrt {{x^4} + 1} }} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}\dfrac{{1\cdot 3\cdot 5\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot 4\cdot 6\cdot\cdot\cdot 2n}}\dfrac{1}{{4n + 1}}$
${S_{125}} \approx 0.926987$
Work Step by Step
Let us expand ${\left( {1 + x} \right)^{ - 1/2}}$ 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 $\left| x \right| \lt 1$.
Thus,
${\left( {1 + x} \right)^{ - 1/2}} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){x^n}$,
converges for $\left| x \right| \lt 1$.
Substituting ${x^4}$ for $x$ in the series above gives
${\left( {1 + {x^4}} \right)^{ - 1/2}} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){x^{4n}}$,
converges for $\left| {{x^4}} \right| \lt 1$, or $\left| x \right| \lt 1$.
Thus, $\dfrac{1}{{\sqrt {{x^4} + 1} }} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){x^{4n}}$.
Taking the definite integral on the series gives
$\mathop \smallint \limits_0^1 \dfrac{{{\rm{d}}x}}{{\sqrt {{x^4} + 1} }} = \mathop \sum \limits_{n = 0}^\infty \mathop \smallint \limits_0^1 \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){x^{4n}}{\rm{d}}x$
$ = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right)\left[ {\dfrac{{{x^{4n + 1}}}}{{4n + 1}}} \right]_0^1 = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right)\dfrac{1}{{4n + 1}}$
From page 585, we know that
$\left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right) = \dfrac{{ - \dfrac{1}{2}\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)\cdot\cdot\cdot\left( { - \dfrac{{2n - 1}}{2}} \right)}}{{1\cdot 2\cdot 3\cdot\cdot\cdot n}} = {\left( { - 1} \right)^n}\dfrac{{1\cdot 3\cdot 5\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot 4\cdot 6\cdot\cdot\cdot 2n}}$
Write
$S = \mathop \smallint \limits_0^1 \dfrac{{{\rm{d}}x}}{{\sqrt {{x^4} + 1} }} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}\dfrac{{1\cdot 3\cdot 5\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot 4\cdot 6\cdot\cdot\cdot 2n}}\dfrac{1}{{4n + 1}}$
The alternating series above has positive terms: ${b_n} = \dfrac{{1\cdot 3\cdot 5\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot 4\cdot 6\cdot\cdot\cdot 2n}}\dfrac{1}{{4n + 1}}$.
By Eq. (2) in Section 11.4:
$\left| {S - {S_n}} \right| \lt {b_{n + 1}}$
We choose $n$ so that
$\left| {S - {S_n}} \right| \lt {b_{n + 1}} \lt {10^{ - 4}}$
$\dfrac{{1\cdot 3\cdot 5\cdot\cdot\cdot\left( {2n + 1} \right)}}{{2\cdot 4\cdot 6\cdot\cdot\cdot\left( {2n + 2} \right)}}\dfrac{1}{{4n + 5}} \lt {10^{ - 4}}$
$\dfrac{{2\cdot 4\cdot 6\cdot\cdot\cdot\left( {2n + 2} \right)}}{{1\cdot 3\cdot 5\cdot\cdot\cdot\left( {2n + 1} \right)}}\left( {4n + 5} \right) \gt {10^4}$
We evaluate and list some values for $n$ in the following table:
$\begin{array}{*{20}{c}}
n&{}&{\dfrac{{2\cdot 4\cdot 6\cdot\cdot\cdot\left( {2n + 2} \right)}}{{1\cdot 3\cdot 5\cdot\cdot\cdot\left( {2n + 1} \right)}}\left( {4n + 5} \right)}\\
{124}&{}&{9938.07}\\
{125}&{}&{10057.3}\\
{126}&{}&{10177.1}\\
{127}&{}&{10297.3}
\end{array}$
From the results in the table above, we choose $N=125$ so that the series value is within an error of at most ${10^{ - 4}}$ to the definite integral.
We compute the following:
${S_{125}} \approx 0.926987$, ${\ \ \ \ \ }$ $S \approx 0.927037$
It is verified that $\left| {S - {S_{125}}} \right| \approx 0.00005 \lt {10^{ - 4}}$.
