is it correct to say $\sum{(-1)^n*\cos(\ln(n)x)}=\operatorname{Re}[(1-2^{-ix})\zeta(1+ix)]$

Question

The statement is that for a>0 $$\sum\frac{(-1)^n\cos(\ln(n)x)}{n^a}=\operatorname{Re}[(1-2^{1-(a+ix)}\zeta(a+ix))]$$ Here is how I got that stament

$$\sum\frac{(-1)^n}{n^s}=1-2^{1-s}\zeta(s))$$ then $$e^{\sum\frac{(-1)^n}{n^s}}=e^{1-2^{1-s}\zeta(s))}$$ $$\prod e^{\sum\frac{(-1)^n}{n^s}}=e^{1-2^{1-s}\zeta(s))}$$ Knowing that To find the product of imaginary numbers you multiply the lengths and add the angles $$|e^{\frac{(-1)^n}{n^s}}|=e^\frac{(-1)^n\cos(\ln(n)x)}{n^a}$$ and that the angle $$\angle(e^{\frac{(-1)^n}{n^s}})=\frac{(-1)^n\sin(\ln(n)x)}{n^a}$$ so the magnitue is $$ e^{\sum\frac{(-1)^n\cos(\ln(n)x)}{n^a}}$$ and the angle is $$\sum\frac{(-1)^n*\sin(\ln(n)x)}{n^a}$$ whitch gives us $$e^{\sum\frac{(-1)^n\cos(\ln(n)x)+i(-1)^n\sin(\ln(n)x)}{n^a}}=e^{1-2^{1-s}\zeta(s))}$$ loging both sides we get $$\sum{\frac{(-1)^n\cos(\ln(n)x)+i(-1)^n\sin(\ln(n)x)}{n^a}}={1-2^{1-s}\zeta(s))}$$ so $$\sum\frac{(-1)^n\cos(\ln(n)x)}{n^a}=\operatorname{Re}[(1-2^{1-(a+ix)}\zeta(a+ix)]$$ By braking up the cos you can make a bunch of other cool statments.

there are also a bunch of other cool statements you can make like $$(1-2^{1-(a+ix)})\zeta(a+ix)+(1-2^{1-(a-ix)})\zeta(a-ix)+i[(1-2^{1-(a+ix)})\zeta(a+ix)-(1-2^{1-(a-ix)})\zeta(a-ix)]=(1-2)^{1-(a+ix)}\zeta(a+ix)$$

This is also super cool if we take $\sum\frac{(-1)^n\cos(\ln(n)x)}{n^a}$ and squire it we get or at least i think we get $$\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{(-1)^n\cos(\ln(n)x)*(-1)^m\cos(\ln(m)x)}{n^a}$$ if you do the same thing to sin and add them you get $$\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{(-1)^n\cos(\ln(n)x)*(-1)^m\cos(\ln(m)x)+(-1)^n\sin(\ln(n)x)*(-1)^m\sin(\ln(m)x)}{n^a}$$ whitch acoring to wolfram alfa is thought some algabric trickery http://www.wolframalpha.com/input/?i=sin(ln(m))*sin(ln(n))%2Bcos(ln(m))*cos(ln(n))

(we know they have the same sin becuse there multipled by the same negitve) $$\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{(-1)^{mn}\cos(\ln(\frac{n}{m})x)}{(mn)^a}$$

with means that $$\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{(-1)^{mn}\cos(\ln(\frac{n}{m})x)}{(mn)^a}=\operatorname{MAG}[(1-2^{1-(a+ix)}\zeta(a+ix))]^2$$

and it just keeps getting cooler we can expencate both sides and get $$e^{\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{(-1)^{mn}\cos(\ln(\frac{n}{m})x)}{(mn)^a}}=e^{\operatorname{MAG}[(1-2^{1-(a+ix)}\zeta(a+ix))]^2}$$ which is $$\prod_{n=0}^\infty e^{\sum_{m=0}^\infty\frac{(-1)^{mn}\cos(\ln(\frac{n}{m})x)}{(mn)^a}}=e^{\operatorname{MAG}[(1-2^{1-(a+ix)}\zeta(a+ix))]^2}$$ which is $$\prod_{n=0}^\infty*\prod_{m=0}^\infty e^{\frac{(-1)^{mn}\cos(\ln(\frac{n}{m})x)}{(mn)^a}}=e^{\operatorname{MAG}[(1-2^{1-(a+ix)}\zeta(a+ix))]^2}$$

Answer 1

Same answer as your previous question.

Trying to avoid complex numbers makes everything messy. Of course if it helps you to understand, good for you, but you should realize complex numbers make everything much cleaner.

For $\Re(s) > 1$, $$\zeta(s) = \sum_{n=1}^\infty n^{-s}, \qquad (1-2^{1-s})\zeta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$$ This last series is analytic for $\Re(s) > 0$ thus by analytic continuation the equality stays true for $\Re(s) > 0$, and for $a > 0, b \in \mathbb{R}$ $$\Re((1-2^{1-a-ib})\zeta(a+ib)) = \sum_{n=1}^\infty (-1)^{n+1} \Re(n^{-a-ib}) = \sum_{n=1}^\infty (-1)^{n+1} n^{-a} \cos(b \ln n)$$ And hence $$\Re((1-2^{1-a-ib})\zeta(a+ib))^2 = \sum_{m=1}^\infty \sum_{n=1}^\infty (-1)^{n+m} (nm)^{-a} \cos(b \ln n) \cos(b \ln m)$$ $$|(1-2^{1-a-ib})\zeta(a+ib)|^2 = (1-2^{1-a-ib})\zeta(a+ib)(1-2^{1-a+ib})\zeta(a-ib)\\=\sum_{m=1}^\infty \sum_{n=1}^\infty (-1)^{n+m} (nm)^{-a} (n/m)^{-ib}=\sum_{m=1}^\infty \sum_{n=1}^\infty (-1)^{n+m} (nm)^{-a} \cos(b \log(n/m))$$ Where, since $\sum_{n=1}^\infty (2n-1)^{-s}-(2n)^{-s}$ converges absolutely, changing the order of summation is allowed provided the terms stay grouped by two