$p$-series divided by alternating $p$-series = geometric series? Why?

Question

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."

The numerator is called the Riemann zeta function and the denominator is called the Dirichlet eta function, so you're asking for $\frac{\zeta(p)}{\eta(p)}$. Try writing out $\zeta(p)-\eta(p)$. — Akiva Weinberger, Apr 27 '15 at 16:12

score 8 · Answer 1 · edited Apr 28 '15 at 08:51

8

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

edited Apr 28 '15 at 08:51

Newb

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answered Apr 27 '15 at 16:00

Elaqqad

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Nice. Adding to the files. – Simon S Apr 27 '15 at 16:01

$p$-series divided by alternating $p$-series = geometric series? Why?

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