Answer
(a) We show that
$\dfrac{1}{{1 + x}} = \mathop \sum \limits_{n = 0}^N {\left( { - 1} \right)^n}{x^n} + \dfrac{{{{\left( { - 1} \right)}^{N + 1}}{x^{N + 1}}}}{{1 + x}}$
(b) We show that
$\ln 2 = \mathop \sum \limits_{n = 1}^N \dfrac{{{{\left( { - 1} \right)}^{n - 1}}}}{n} + {\left( { - 1} \right)^{N + 1}}\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$
(c) We show that as $N \to \infty $, $\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$ approaches zero.
(d) We prove:
$\ln 2 = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdot\cdot\cdot$
Work Step by Step
(a) From Table 2, we have
$\dfrac{1}{{1 + x}} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{x^n} = 1 - x + {x^2} - {x^3} + {x^4} - \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \lt 1$
Write as
(1) ${\ \ \ \ \ }$ $\dfrac{1}{{1 + x}} = \mathop \sum \limits_{n = 0}^N {\left( { - 1} \right)^n}{x^n} + \mathop \sum \limits_{n = N + 1}^\infty {\left( { - 1} \right)^n}{x^n}$
Notice that the second sum on the right-hand side of equation (1) can be written as
$\mathop \sum \limits_{n = N + 1}^\infty {\left( { - 1} \right)^n}{x^n} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^{N + 1 + n}}{x^{N + 1 + n}}$
Thus,
$\mathop \sum \limits_{n = N + 1}^\infty {\left( { - 1} \right)^n}{x^n} = {\left( { - 1} \right)^{N + 1}}{x^{N + 1}}\mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{x^n}$
Since $\mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{x^n} = \dfrac{1}{{1 + x}}$, so $\mathop \sum \limits_{n = N + 1}^\infty {\left( { - 1} \right)^n}{x^n} = \dfrac{{{{\left( { - 1} \right)}^{N + 1}}{x^{N + 1}}}}{{1 + x}}$.
Therefore, equation (1) becomes
$\dfrac{1}{{1 + x}} = \mathop \sum \limits_{n = 0}^N {\left( { - 1} \right)^n}{x^n} + \dfrac{{{{\left( { - 1} \right)}^{N + 1}}{x^{N + 1}}}}{{1 + x}}$
(b) From part (a), we obtain
$\dfrac{1}{{1 + x}} = \mathop \sum \limits_{n = 0}^N {\left( { - 1} \right)^n}{x^n} + \dfrac{{{{\left( { - 1} \right)}^{N + 1}}{x^{N + 1}}}}{{1 + x}}$
Integrating from $0$ to $1$:
$\mathop \smallint \limits_0^1 \dfrac{1}{{1 + x}}{\rm{d}}x = \mathop \sum \limits_{n = 0}^N {\left( { - 1} \right)^n}\mathop \smallint \limits_0^1 {x^n}{\rm{d}}x + {\left( { - 1} \right)^{N + 1}}\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$
$\left[ {\ln \left( {1 + x} \right)} \right]_0^1 = \mathop \sum \limits_{n = 0}^N {\left( { - 1} \right)^n}\dfrac{1}{{n + 1}}\left( {{x^{n + 1}}} \right)|_0^1 + {\left( { - 1} \right)^{N + 1}}\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$
$\ln 2 = \mathop \sum \limits_{n = 0}^N \dfrac{{{{\left( { - 1} \right)}^n}}}{{n + 1}} + {\left( { - 1} \right)^{N + 1}}\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$
By shifting the index, we get
$\ln 2 = \mathop \sum \limits_{n = 1}^N \dfrac{{{{\left( { - 1} \right)}^{n - 1}}}}{n} + {\left( { - 1} \right)^{N + 1}}\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$
(c) From part (b), we obtain
$\ln 2 = \mathop \sum \limits_{n = 1}^N \dfrac{{{{\left( { - 1} \right)}^{n - 1}}}}{n} + {\left( { - 1} \right)^{N + 1}}\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$
Consider the integrand $\dfrac{{{x^{N + 1}}}}{{1 + x}}$ on the right-hand side:
For $0 \lt x \lt 1$, we have
$0 \lt \dfrac{{{x^{N + 1}}}}{{1 + x}} \lt {x^{N + 1}}$
Therefore,
$0 \lt \mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x \lt \mathop \smallint \limits_0^1 {x^{N + 1}}{\rm{d}}x = \dfrac{1}{{N + 1}}\left( {{x^{N + 1}}} \right)|_0^1$
$0 \lt \mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x \lt \dfrac{1}{{N + 1}}$
Taking the limit as $N \to \infty $,
$0 \lt \mathop {\lim }\limits_{N \to \infty } \mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x \lt \mathop {\lim }\limits_{N \to \infty } \dfrac{1}{{N + 1}}$
$0 \lt \mathop {\lim }\limits_{N \to \infty } \mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x \lt 0$
By Squeeze Theorem: as $N \to \infty $, $\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$ approaches zero.
(d) From part (b), we obtain
$\ln 2 = \mathop \sum \limits_{n = 1}^N \dfrac{{{{\left( { - 1} \right)}^{n - 1}}}}{n} + {\left( { - 1} \right)^{N + 1}}\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$
Also from part (c), we conclude that as $N \to \infty $, $\mathop \smallint \limits_0^1 \dfrac{{{x^{N + 1}}}}{{1 + x}}{\rm{d}}x$ approaches zero.
Therefore, as $N \to \infty $, we obtain
$\ln 2 = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}$
Hence,
$\ln 2 = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdot\cdot\cdot$
