I'm having some troubles in understanding why a REAL sum turns out to be COMPLEX.
The sum in question is the following:
$$\sum_{k = 1}^{+\infty} (-1)^{k+1} b(k) \omega^{-k/4} \sin\left(\frac{k\pi}{8}\right)$$
Where
$$b(k) = \frac{\Gamma\left(\frac{k+7}{4}\right)}{\Gamma\left(\frac{k+4}{4}\right)\left(\frac{k+3}{4}\right)}$$
Now, as you might see, nothing complex or imaginary occurs, also because I'm studying the case
$$\boxed{\omega >0}$$
Since $k$ runs around naturals, and $b(k)$ is a ratio between positive reals, and the sine function cannot get complex or imaginary for those values of $k$, how is possible that the sum turns up to be complex?
I used W. Mathematica (serious version, not Alpha) and one of the terms (writing down it all would be a hell) is the following:
$$-\frac{2 \color{red}{(-1)^{3/8}} \left(\color{red}{\sqrt[4]{-1}}-1\right) \, _3F_2\left(\frac{1}{2},1,1;\frac{5}{8},\frac{9}{8};-\frac{1}{\omega ^2}\right)}{\sqrt[4]{\omega } \Gamma \left(\frac{1}{4}\right)}$$
How do those RED terms come out?
Thank you in advance
Just because an expression has complex numbers in it, doesn't mean it's actually complex. Expanding out gives $$ (-1)^{3/8}(\sqrt[4]{-1}-1) = (-1)^{5/8}-(-1)^{3/8} = e^{5\pi i/8} - e^{3\pi i/8} \\ = e^{\pi i/2} (e^{i\pi/8}-e^{-i\pi/8}) = i \cdot 2i\sin{\left(\frac{\pi}{8}\right)} = -2\sin{\left(\frac{\pi}{8}\right)}. $$ This is real, of course; one finds using the half-angle formulae that $$ \sin{\left(\frac{\pi}{8}\right)} = \sqrt{\frac{1-\cos{(\pi/4)}}{2}} = \sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}} = \frac{1}{2}\sqrt{2-\sqrt{2}}. $$ (And in case you're wondering, Mathematica agrees if you use
ExpToTrig[Expand[]]on it.)