Answer
(a) See confirmation below.
(b) See the result below.
(c) $n=50$.
Work Step by Step
(a) The series can be written as $S = \mathop \sum \limits_{k = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{k - 1}}}}{{2k - 1}}$, where ${a_k} = \dfrac{1}{{2k - 1}}$.
Using a CAS, we compute the sum and confirm the result:
$S = \mathop \sum \limits_{k = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{k - 1}}}}{{2k - 1}} = \dfrac{\pi }{4}$
Hence, $\dfrac{\pi }{4} = 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \cdot\cdot\cdot$.
(b) Using a CAS we compute ${s_{25}}$ and $\left| {\dfrac{\pi }{4} - {s_{25}}} \right|$:
${s_{25}} = \dfrac{{77030060483083029083}}{{96845140757687397075}} \approx 0.795394$
$\left| {\dfrac{\pi }{4} - {s_{25}}} \right| \approx 0.00999601$
Hence, $\left| {\dfrac{\pi }{4} - {s_{25}}} \right| \lt {10^{ - 2}}$.
(c) We find the value of $n$ using part (b) of Theorem 9.6.2:
$\left| {S - {s_n}} \right| \le {a_{n + 1}} \lt {10^{ - 2}}$
$\dfrac{1}{{2n + 1}} \lt {10^{ - 2}}$
$2n + 1 \gt 100$
$n \gt 49.5$
The required value of $n$ is $n=50$.
