About solving an equation Let us consider an equation of the form 
$$\sum_{n=1}^{\infty}\frac{(-1)^n\sin(y\ln n)}{n^{x}}=-\frac{y}{a}$$
My question is:
Is it possible to say that $y=0$ is the only solution of this equation.
 A: For large $x$ ($x=10$ on the plot), the LHS divided by $y$ is very similar to a $\text{sinc}$ so that for large $a$ there can probably be numerous solutions.

(For smaller $x$, I guess that the behavior is oscillating anyway and numerous roots remain possible.)
A: The sum can be written in terms of the Riemann zeta function:
$$\begin{align}
 F(x+iy)=\sum_{n=1}^{\infty}\frac{(-1)^n\sin(y\ln n)}{n^{x}}&=\Im \left((1-2^{1-(x+iy)})\zeta(x+iy)\right) .\end{align}
$$
This is because $\sin (y\ln n) = -\Im n^{ -iy}$. The series converges if $x>0$ and the formula holds by the analytic continuation of the Riemann zeta function. 
Let $\rho=1/2+it_0$ with $t_0\approx14.134725$, the imaginary part of the first  nontrivial zero of zeta function. Then by the open mapping theorem, $F(\tau)\neq 0$ for some $\tau=x+iy\in\mathbb{C}$ with $|\tau-\rho|<0.1$.  Then we find $a$ such that $F(\tau)=-\frac ya$. This $y$ is a solution to the desired equation. 
A: I am not sure it's the right answer. You can find a pair $(x,y)$ that satisfies the equation. For example, you can choose $x=2$ and then solve $$\sum_{n=1}^{\infty}\frac{(-1)^n\sin(y\ln n)}{n^{2}}=-\frac{y}{a}$$
