I was solving a series based question and I am now totally confused on how to proceed further to find the analytical values of $a$ and $b$ (where $a,b\in \mathbb{R}$) such that:
$$ \sum_{k=2}^{\infty}\frac{(-1)^k\cos(b \ln(k))}{k^a} = 1 $$
and
$$ \sum_{k=2}^{\infty}\frac{(-1)^k\sin(b \ln(k))}{k^a} = 0 $$
where $\ln x$ is the natural logarithm.
any help would be greatly appreciated. Thanks in advance :)
We begin with the equations in the posted question
$$\sum_{n=2}^\infty \frac{(-1)^n\sin(b\log(n))}{n^a}=0\tag1$$
and
$$\sum_{n=2}^\infty \frac{(-1)^n\cos(b\log(n))}{n^a}=1\tag2$$
Multiplying $(1)$ by $i$ and subtracting from $(2)$, we find that the pair of equations $(1)$ and $(2)$ are equivalent to the equation
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^z}=0\tag3$$
where $z=a+ib$.
The series on the left-hand side of $(3)$ is a representation of the Dirichlet Eta Function, $\eta(z)$, for $\text{Re}(z)>0$. Therefore, the posted question implicitly asks to find the zeros of the Eta function in the right-half plane.
To proceed, we note the relationship
$$\underbrace{\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^z}}_{\eta(z), \text{Re}(z)>0}=(1-2^{1-z})\underbrace{\sum_{n=1}^\infty \frac1{n^z}}_{\zeta(z),\text{Re}(z)>1}\tag4$$
where the series on the right-hand side of $(4)$ is a representation of the Riemann Zeta function, $\zeta(z)$, for $\text{Re}(z)>1$.
ASIDE:
- I showed in THIS ANSWER and THIS ANSWER that the series representation for $\zeta(z)$ in $(4)$ diverges for $\text{Re}(z)\le 1$.
- We can extend the definition of $\zeta(z)$ to the right-half plane by rewriting $(4)$ as
$$\zeta(z)=(1-2^{1-z})^{-1}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^z}\tag5$$
for $\text{Re}(z)>0$. I showed in THIS ANSWER that from $(5)$ we can write
$$\zeta(z)=\frac1{z-1}+\gamma +O(z-1)$$
which explicitly shows that $\zeta(z)$ has a simple pole at $z=1$ with residue $1$.
It is easy to see from $(5)$ that zeros of $1-2^{1-z}$ are also zeros of $\eta(z)$ unless they are also poles of $\zeta(z)$. Therefore, we find that the zeros of $\eta(z)$, as represented by the series in $(3)$, include the points $z_k$ where
$$z_k=1+i\frac{2\pi k}{\log(2)}$$
for $k\in \mathbb{Z}\setminus \{0\}$.
Inasmuch as the zeros of the Zeta function lie in the region $0<\text{Re}(z)<1$, the Dirichlet Eta function has no zeros on the half plane $\text{Re}(z)>1$. It had been conjectured (The Riemann Hypothesis, 1859) that all of the zeros of $\zeta(z)$ are located at even negative integers (the trivial zeros) and at complex numbers with real part equal to $1/2$. This conjecture remains unproven at the time of this post. If this conjecture is true, then all other zeros of $\eta(z)$ as represented by the series in $(3)$ are located at points $z=\frac12+ib$.
ZEROS OF THE ETA FUNCTION
The Eta function can be analytically continued to the left-half plane through the relationship in $(5)$ (i.e., $\eta(z)=(1-2^{1-z})\zeta(z)$) with the Zeta function and the analytic continuation of the Riemann Zeta function through the functional equation
$$\zeta(z)=2^z\pi^{z-1}\sin\left(\frac{\pi z}{2}\right)\Gamma(1-z)\zeta(1-z)\tag6$$
Note from $(6)$ that $\zeta(z)$ has zeros at the negative even integers. These are the so-called trivial zeros of $\zeta(z)$. So, we see that the zeros of $\eta(z)$ include the points
$$z_k=\begin{cases}1+i\frac{2k\pi}{\log(2)}&,k\in \mathbb{Z}\setminus \{0\}\\\\ 2k&,k\in \mathbb{Z_{<0}} \end{cases}$$
Again, if the Riemann hypothesis is correct, all other zeros of $\eta(z)$ are located at points $z=\frac12+ib$.
