Maclaurin expansion for $f(x) = \log \frac{1+x}{1-x}$. Taking $x = 1/3$ how many terms do we need to approximate the value of $\log 2$ with $6$ exact decimal places?
I've tried to do the Maclaurin expansion, taking into account $\log$ is the same as $\ln$, but in the second derivate I have an indetermination $0/n$ and I don't know how to continue.
We have
$$ f(x) = \log \frac{1+x}{1-x} = \sum_{n = 0}^{t}\frac{2x^{2n+1}}{2n+1} + \sum_{k = t+1}^{\infty}\frac{2x^{2k+1}}{2k+1} < \sum_{n = 1}^{t}\frac{2x^{2n+1}}{2n+1} + \frac{2}{2t+3}\sum_{k = t+1}^{\infty}x^{2k+1} $$
$$ \log \frac{1+x}{1-x} < \sum_{n = 1}^{t}\frac{2x^{2n+1}}{2n+1} + \frac{2}{2t+3} \cdot\frac{x^{2t+3}}{1 - x^2} $$
For $x = 1/3$, to get $f(1/3) = \log 2$ accurate to six decimal places, we need the minimum value of $t$ for which $$ \frac{2}{2t+3} \cdot\frac{(1/3)^{2t+3}}{1 - (1/3)^2} < 10^{-6} $$ The smallest value of $t$ for which this happens is $t = 5$. Hence we need a minimum of $5$ terms to get $\log 2$ accurate to six decimal places.