# How do you solve this question?

May 15, 2018

See below

#### Explanation:

(1) $f \left(x\right) = \ln \left(x\right) \implies f \left(1\right) = 0$

${f}^{'} \left(x\right) = \frac{1}{x} \implies {f}^{'} \left(1\right) = 1$
$f ' ' \left(x\right) = - \frac{1}{x} ^ 2 \implies f ' ' \left(1\right) = - 1$
$f ' ' ' \left(x\right) = \frac{\left(- 1\right) \left(- 2\right)}{x} ^ 3 \implies f ' ' ' \left(1\right) = 2$
${f}^{\left(4\right)} \left(x\right) = \frac{\left(- 1\right) \left(- 2\right) \left(- 3\right)}{x} ^ 4 \implies {f}^{\left(4\right)} \left(1\right) = - 6$

The general pattern should be clear from this:

${f}^{\left(n\right)} \left(x\right) = \frac{\left(- 1\right) \left(- 2\right) \ldots \left(- n + 1\right)}{x} ^ n \implies$
 f^((n))(1) =(-1)^(n-1)(n-1)!

(2) So, the Taylor series for $\ln \left(x\right)$ centered at $x = 1$ is given by

ln (x) = sum_{n=0}^oo f^((n))(1)/(n!) (x-1)^n
qquadqquad = sum_{n=0}^oo (-1)^(n-1)((n-1)!)/(n!) (x-1)^n
$q \quad q \quad = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n - 1} {\left(x - 1\right)}^{n} / n$

(3) The $n$-th term in the Taylor series above is given by

${t}_{n} = {\left(- 1\right)}^{n - 1} {\left(x - 1\right)}^{n} / n$

and thus

${t}_{n + 1} / {t}_{n} = - \frac{n}{n + 1} \left(x - 1\right)$

For the series to converge, we must have

${\lim}_{n \to \infty} | {t}_{n + 1} / {t}_{n} | < 1 \implies$

${\lim}_{n \to \infty} | - \frac{n}{n + 1} \left(x - 1\right) | < 1 \implies$

$\textcolor{red}{| x - 1 | < 1}$
which is the domain of convergence.

(4)

$\ln \left(\frac{4}{3}\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n - 1} {\left(\frac{4}{3} - 1\right)}^{n} / n$
$q \quad q \quad = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n - 1} \frac{1}{n {3}^{n}}$

$\ln \left(\frac{2}{3}\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n - 1} {\left(\frac{2}{3} - 1\right)}^{n} / n$
$q \quad q \quad = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n - 1} {\left(- 1\right)}^{n} / \left(n {3}^{n}\right)$

Hence

$\ln \left(2\right) = \ln \left(\frac{4}{3}\right) - \ln \left(\frac{2}{3}\right)$
$q \quad = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n - 1} \frac{1 - {\left(- 1\right)}^{n}}{n {3}^{n}}$

In this sum, all the even terms cancel, so that we get

$\ln \left(2\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{2 n + 1 - 1} \frac{2}{\left(2 n + 1\right) {3}^{2 n + 1}}$

and hence

$\ln \left(\sqrt{2}\right) = \frac{1}{2} \ln \left(2\right) = {\sum}_{n = 0}^{\infty} \frac{1}{\left(2 n + 1\right) {3}^{2 n + 1}}$