TITLE: $p$-series divided by alternating $p$-series = geometric series? Why?
QUESTION [7 upvotes]: I thought the following equation was interesting:
$\dfrac{1 + \frac{1}{2^p} + \frac{1}{3^p} + ...  }{1 - \frac{1}{2^p} + \frac{1}{3^p} - ...} = \dfrac{1}{1-2^{1-p}}$ for $p>1$, where $p$ is a real number.
So in other words, this ratio on the LHS is actually a geometric series in disguise.
I can derive the expression above by doing some term rearrangements, but I'm curious...

is there a slick way to see that this ratio is on the LHS is a geometric series with $r = 2^{1-p}$?
is there part of some thing more general? (ratio of one series over its alternating version equals a geometric series?)


Edit: I also would love to see other solutions for showing that equality. I like my solution but it feels very "high school math."

REPLY [8 votes]: Let:
$$x=\sum_{i=1}^\infty \frac{1}{i^p}\ \ \ \ \ \ y=\sum_{i=1}^{\infty}\frac{(-1)^{i +1}}{i^p} $$
We have:
$$x-y=2\times\frac{1}{2^p}x $$