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I am trying to estimate $ln(\frac{3}{2})$ to three decimal places. I was trying to use Taylor series for $ln(x+1)$ with Lagrange's form of reminder. As

$$(ln(x+1))^{(n)} = \frac{(-1)^{n+1}(n-1)!}{(x+1)^{n}}$$

the n-th reminder has the following form:

$$r_n(x) = \frac{(-1)^{n}}{(c+1)^{n}}x^{n}$$

In our case $c\in (0,\frac{1}{2})$, so we substitute c = 0 as we want to limit from the top. Putting $\frac{1}{2}$ as x we get

$$|r_n(\frac{1}{2})| < \frac{1}{n2^n}$$ To make sure $|r_n(1/2)| < 0,001$ we solve $\frac{1}{n2^n} < 0,001$ which gives n = 8, $r_8(\frac{1}{2}) \approx 0.000488281$. That would mean that I would have to compute seven elements of Taylor polynomial at x = $\frac{1}{2}$, which seems like a rather tedious task. This brings me to my question: are my computations correct?

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You are correct exploiting tha alternate series. However, you can make it much faster using $$\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n$$ $$\log(1-x)=-\sum_{n=1}^\infty \frac{1}{n}x^n$$ $$\log \left(\frac{1+x}{1-x}\right)=2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}$$ Now $\frac{1+x}{1-x}=\frac 32 \implies x=\frac 15$

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