Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

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

share|improve this question
3  
If $x+iy$ is a nontrivial zero of Riemann $\zeta$, sure... –  J. M. Aug 10 '11 at 8:01
1  
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

1 Answer 1

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 $\eta$.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.