Example 3. Expand the function $f(x)=\ln \frac{1+x}{1-x}$ into a Maclaurin series and, using this expansion, compute $\ln 3$ with an accuracy of $10^{-3}$.
Solution. We will use the expansion from part b) of Example 1:

$$
\frac{1}{1-t}=1+t+t^{2}+t^{3}+\ldots+t^{n}+\ldots
$$

Replace \( t \) with \( t^2 \) in this expansion. We get

$$
\frac{1}{1-t^{2}}=1+t^{2}+t^{4}+t^{6}+\ldots+t^{2 n}+\ldots \quad, \quad(|t|<1)
$$

Integrate this equality over the interval \([0 ; x]\) for \( x \in (-1 ; 1) \) (see Theorem 17). On one side, the integral of the left part is \(\int_{0}^{x} \frac{d t}{1-t^{2}} = \left.\frac{1}{2} \ln \left|\frac{1+t}{1-t}\right|\right|_{0} ^{x} = \frac{1}{2} \ln \frac{1+x}{1-x}, \quad |x|<1\).

On the other side, term-by-term integration of the right part of this equality leads to

$$
\begin{aligned}
\left(t+\frac{t^{3}}{3}+\frac{t^{5}}{5}+\ldots+\frac{t^{2 n+1}}{2 n+1}+\ldots\right)\left.\right|_{0} ^{x}= \\
& =x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots+\frac{x^{2 n+1}}{2 n+1}+\ldots \quad(|x|<1)
\end{aligned}
$$

By equating the obtained results and multiplying by 2, we arrive at the required expansion:

$$
\ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots+\frac{x^{2 n+1}}{2 n+1}+\ldots\right) \quad(|x|<1)
$$

To compute \(\ln 3\), it is sufficient to set \( x = \frac{1}{2} \) in this equality. We have \(\ln 3 = 2\left(\frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2^{3}} + \frac{1}{5} \cdot \frac{1}{2^{5}} + \frac{1}{7} \cdot \frac{1}{2^{7}} + \ldots\right)\).

To achieve a precision of the order of \(10^{-3}\), it is sufficient to take such \( n \) that \( r_{n} < 10^{-3} \). Using the equality \((|t|<1) 1+t+t^{2}+\ldots=\frac{1}{1-t}\), we have \(\frac{1}{9} \cdot \frac{1}{2^{9}} + \frac{1}{11 \cdot 2^{11}} + \ldots < \frac{1}{9} \cdot \frac{1}{2^{9}} \left(1 + \frac{1}{2^{2}} + \frac{1}{2^{4}} + \ldots\right) = \frac{1}{9 \cdot 2^{9}} \cdot \frac{1}{1-\frac{1}{4}} = \frac{1}{9 \cdot 2^{9}} \cdot \frac{4}{3} = \frac{4}{9 \cdot 2^{9} \cdot 3} = \frac{1}{27 \cdot 2^{7}} = \frac{1}{3456} < 10^{-3}\).

Thus, \(\ln 3 \approx 2(0.5 + 0.0417 + 0.0062 + 0.0011) = 1.098\).

Answer. \(\ln 3 = 1.098 \pm 10^{-3}\).