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$\sum_{n=1}^{\infty}\frac{\left ( -1 \right )^n}{n^x}\cos{\left ( y\ln{n} \right )}$ $\sum_{n=1}^{\infty}\frac{\left ( -1 \right )^n}{n^x}\sin{\left ( y\ln{n} \right )}$

$x$ and $y$ are arbitrary real number, and $x>0$.

Question. Is it possible that these series's value are $0$?

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    Scratch that; if $x+iy$ is a (trivial or nontrivial) zero of Riemann's famed function, then both series should be zero. Your series are the real and imaginary parts of [$-\eta(x-iy)$](http://mathworld.wolfram.com/DirichletEtaFunction.html).2011-08-10

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To settle this question:

The two series in the question are respectively the real and imaginary parts of $-\eta(x-iy)$, where $\eta(s)$ is the Dirichlet $\eta$ function. Thus for real $x$ and $y$, if $x+iy$ is a nontrivial zero (recall that the series converge only for x > 0) of the Riemann $\zeta$ function, both series will be zero. Additionally, since $\eta(s)=(1-2^{1-s})\zeta(s)$, $x=1$ and $y=\frac{2\pi i k}{\ln\,2}$ with $k$ a nonzero integer would also be zeroes. For the analytically continued Dirichlet $\eta$ function, the "trivial" zeroes of Riemann $\zeta$ will also be zeroes of Dirichlet $\eta$.

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    Oh yes, let me add that in... thanks @anon.2011-08-11