Answer
$F\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)\left( {2n + 1} \right)!}} = x - \dfrac{{{x^3}}}{{3\cdot3!}} + \dfrac{{{x^5}}}{{5\cdot5!}} - \dfrac{{{x^7}}}{{7\cdot7!}} + \cdot\cdot\cdot$
$F\left( 1 \right)$ to three decimal places is ${F_2}\left( 1 \right) \approx 0.946111$.
Work Step by Step
From Table 2:
$\sin x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}} = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
So,
$\dfrac{{\sin t}}{t} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n}}}}{{\left( {2n + 1} \right)!}} = 1 - \dfrac{{{t^2}}}{{3!}} + \dfrac{{{t^4}}}{{5!}} - \dfrac{{{t^6}}}{{7!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $t$
Taking the definite integral on the series above gives
$F\left( x \right) = \mathop \smallint \limits_0^x \dfrac{{\sin t}}{t}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \mathop \smallint \limits_0^x \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n}}}}{{\left( {2n + 1} \right)!}}{\rm{d}}t$
$ = \mathop \sum \limits_{n = 0}^\infty \left[ {\dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 1}}}}{{\left( {2n + 1} \right)\left( {2n + 1} \right)!}}} \right]_0^x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)\left( {2n + 1} \right)!}}$
Thus,
$F\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)\left( {2n + 1} \right)!}} = x - \dfrac{{{x^3}}}{{3\cdot3!}} + \dfrac{{{x^5}}}{{5\cdot5!}} - \dfrac{{{x^7}}}{{7\cdot7!}} + \cdot\cdot\cdot$
For $x=1$, we have $F\left( 1 \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right)\left( {2n + 1} \right)!}}$, where the positive terms are ${b_n} = \dfrac{1}{{\left( {2n + 1} \right)\left( {2n + 1} \right)!}}$.
By Eq. (2) in Section 11.4:
$\left| {F\left( 1 \right) - {F_n}\left( 1 \right)} \right| \lt {b_{n + 1}}$
We choose $n$ so that
$\left| {F\left( 1 \right) - {F_n}\left( 1 \right)} \right| \lt {b_{n + 1}} \lt {10^{ - 3}}$
$\dfrac{1}{{\left( {2n + 3} \right)\left( {2n + 3} \right)!}} \lt {10^{ - 3}}$
$\left( {2n + 3} \right)\left( {2n + 3} \right)! \gt {10^3}$
We evaluate and list some values for $n$ in the following table:
$\begin{array}{*{20}{c}}
n&{}&{\left( {2n + 3} \right)\left( {2n + 3} \right)!}\\
1&{}&{600}\\
2&{}&{35280}\\
3&{}&{3265920}
\end{array}$
Using the results in the table above, we choose $N=2$ such that the approximation of the integral for $x=1$ to three decimal places.
We verify the result by computing:
${F_2}\left( 1 \right) \approx 0.946111$, ${\ \ \ \ \ }$ $F\left( 1 \right) \approx 0.946083$
Hence, $F\left( 1 \right)$ to three decimal places is ${F_2}\left( 1 \right) \approx 0.946111$.
