<s>[INST] expand $ \arctan\left(\frac{3x+2}{3x-2}\right)$ into pwer series, find radius of convergence (check solution) I would be grateful if someone could check what I've worked out:
$$ f(x)=\arctan\left(\frac{3x+2}{3x-2}\right)\implies f'(x)=\frac{1}{1+(\frac{3x+2}{3x-2})^2}\cdot \frac{3(3x-2)-3(3x+2)}{(3x-2)^2}$$ 
$$=\frac{(3x-2)^2}{(3x-2)^2+3x+2)^2}\cdot \frac{-12}{(3x-2)^2}=-\frac{3}{2}\cdot \frac{1}{1+(\frac{3}{2}x)^2}$$ 
$$=-\frac{3}{2} \cdot \frac{1}{1+(\frac{3}{2}x)^2}=-\frac{3}{2} \sum_{k=0}^{\infty}\left(-\frac{3}{2}x^2\right)^k $$ 
Which implies $$f(x)=\int f'(x)dx=-\frac{3}{2} \sum_{k=0}^{\infty}\left(-\frac{3}{2} \frac{x^{2k+1}}{2k+1}\right)$$ 
Radius of convergence:
$$\Big|(\frac{3}{2}x)^2\Big|<1 \Rightarrow -\frac{2}{3}<x<\frac{2}{3}$$
Is this correct? Thank you in advance
 [/INST] Here is what you are missing,
$$ f(x)=\int f'(x)dx=-\frac{3}{2} \sum_{k=0}^{\infty}\left((-1)^k\frac{3}{2} \frac{x^{2k+1}}{2k+1}\right) +C $$
$$ \implies f(0) = \arctan(-1) = 0 + C \implies C=-\frac{\pi}{4}. $$
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