<s>[INST] doubt about the calculation of the rest for a taylor's series I'm trying to calculate some numbers with the help of series.
My problem is the calculation of the rest of the series.
For example I have to calculate $log 2$.
I can consider the function $$log (\frac{1+x}{1-x})$$ in which I can put $x= \frac{1}{3}$.
The series of Taylor for this function is:
$$log (\frac{1+x}{1-x}) = 2*\sum_{k=0}^ \infty \frac{x^{2k+1}}{2k+1}$$
$$log (\frac{1+x}{1-x}) = 2*\sum_{k=0}^ n \frac{x^{2k+1}}{2k+1}+ 2*\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} = 2*\sum_{k=0}^ n \frac{x^{2k+1}}{2k+1}+ R_n$$
The passage that I can't understand it this:
$$R_n = 2*\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} < \frac{2 |x|^{2n+3}}{2n+3}\sum_{h=0}^ \infty x^{2k}$$ and then it becomes $\frac{2 |x|^{2n+3}}{2n+3} \frac{1}{1-x^2}$
NOTE
$$R_n = 2*\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} = 2*\sum_{h=0}^ \infty \frac{x^{2h+3}}{2h+3} =2 x^3 \sum_{h=0}^ \infty \frac{(x^2)^h}{2h+3}$$
 [/INST] For $k \ge n+1$, $\vert x \vert^{2k+1} = \vert x \vert^{2n+3} x^{2(k-n-1)}$ and $\frac{1}{2k+1} \le \frac{1}{2n+3}$. Hence
$$2\sum_{k=n+1}^ \infty \frac{x^{2k+1}}{2k+1} \le \frac{2 |x|^{2n+3}}{2n+3}\sum_{h=0}^ \infty x^{2h}.$$
Finally $\sum_{h=0}^ \infty x^{2h}$ is the sum of the terms of a geometric sequence and therefore
$$\sum_{h=0}^ \infty x^{2k} = \frac{1}{1-x^2}$$ which provides the desired result.
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