I noticed that the power series for $\arctan$ is the alternating series of that for $\operatorname{arctanh}$.
Does it have a special meaning or even some kind of special importance?
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You should try to make the body of the Question as self-contained as possible, not relying on the title alone to carry the burden of stating the problem. Mathematical expressions can be posted here using MathJax and $\LaTeX$ markup. – hardmath Feb 18 '16 at 11:10
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The relations between trigonometric and hyperbolic functions are well-known. See Euler's formula. – Lucian Feb 18 '16 at 18:49
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The hyperbolic and trigonometric functions are real/imaginary counterparts of each other.
$$\arctan(ix)=i\,\text{artanh}(x),\\ \text{artanh}(ix)=i\,\arctan(x).$$
For an odd series,
$$\sum_k a_k(ix)^{2k+1}=i\sum_k a_k(-1)^kx^{2k+1}.$$
So yes, there is a fundamental relation.
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It should apply to more than just arctan and arctanh, i.e., all trig functions and hyperbolic functions related by imaginary transformations of the arguments. Incomplete list: https://en.wikipedia.org/wiki/Hyperbolic_function#Hyperbolic_functions_for_complex_numbers – Novice C Feb 18 '16 at 09:35
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@NoviceC: it applies to all odd or even power series. And you let an ordinary series alternate just by changing the sign of the argument. – Feb 18 '16 at 09:38
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An other way to see that is to recall that one easily know that
$$ \arctan'(x) = \frac{1}{1+x^2} = \sum_{n=0}^{+\infty} (-1)^n(x^2)^n, $$
and
$$ \text{artanh}'(x) = \frac{1}{1-x^2} = \sum_{n=0}^{+\infty} (x^2)^n. $$
Integrating these series (for more details, see here) and using that $\arctan(0)=\text{artanh}(0)$, it explains the alternated signs.
