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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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If $x+iy$ is a nontrivial zero of Riemann $\zeta$, sure... – J. M. Aug 10 '11 at 8:01
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 $-\eta(x-iy)$. – J. M. Aug 10 '11 at 8:16
up vote 5 down vote accepted

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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This is a bit incomplete. For $\mathrm{Re}(s)>0$, all zeros of $\eta(s)$ will either be a nontrivial zero of $\zeta(s)$, or will be of the form $1+2\pi i n /\ln 2, n\in\mathbb{Z}-\{0\}$. – anon Aug 11 '11 at 4:01
Oh yes, let me add that in... thanks @anon. – J. M. Aug 11 '11 at 4:02

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