Coincidence that series of arctan is alternating series of artanh?

Question

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?

Answer 1

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.

Answer 2

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.