# oft-worn U(1) modification paths of series and products

So I’ve been playing around with products of the gamma function like so:

$\prod_{n=1}^{\infty}\frac{\Gamma(\frac{4n-3}{4n-2})}{\Gamma(\frac{4n-1}{4n})}=\frac{\Gamma(1/2)}{\Gamma(2/3)}\frac{\Gamma(5/6)}{\Gamma(7/8)}\frac{\Gamma(9/10)}{\Gamma(11/12)}\frac{\Gamma(13/14)}\ldots$

When I notice that swapping half of the terms with their reciprocals is akin to taking the general term to the power $(-1)^{n}$

$\prod_{n=1}^{\infty}\left[\frac{\Gamma(\frac{4n-3}{4n-2})}{\Gamma(\frac{4n-1}{4n})}\right]^{(-1)^{n+1}}=\frac{\Gamma(1/2)}{\Gamma(5/6)}\frac{\Gamma(2/3)}{\Gamma(7/8)}\frac{\Gamma(9/10)}{\Gamma(15/16)}\frac{\Gamma(11/12)}\ldots$

And then it occurs to me: well, what would happen if I just changed that $(-1)^{n}$ term with some power of a root of unity: $e^{2\pi i n/v$, and one little bit of lore about infinite series that I suppose I’ve been curious about for a long time and haven’t actually seen written up anywhere comes to fore:

$A = \sum_{n=0}^{\infty} a_{n}$
well, we can multiply each term by $(-1)^{n}$ yielding:
$B = \sum_{n=0}^{\infty} (-1)^{n}a_{n}$
and then by a root of unity:
$C = \sum_{n=0}^{\infty} e^{2\pi i n/v}a_{n}$

Analogously, if we have a product:
$D = \prod_{n=0}^{\infty} \frac{a_{n}}{b_{n}}$
we can do the above trick as well, giving us:
$E = \prod_{n=0}^{\infty} \left[\frac{a_{n}}{b_{n}}\right]^{(-1)^{n}}$
$F = \prod_{n=0}^{\infty} \left[\frac{a_{n}}{b_{n}}\right]^{e^{2\pi i n/v}}$

The idea here is that there are predictable things that occur when we make the above modifications to series and products, and it would be nice to know (and I’m sure I’ll look into it more closely later, this post is just so that I get my ideas written down) what happens as a result.