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Does anyone know about this series? Looks like it converges depending on the value of $a$. Do you know any closed form solution to find the $a$ that makes the series become zero. $$\sum_{n=1}^\infty \frac{\cos( a \ln(n))}{\sqrt n}$$

Mahdi Rouholamini
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    What makes you sure that there exists at least one value of $a$ for which this converges? – gt6989b Nov 29 '19 at 06:54
  • I read that somewhere on a webpage. I remember there were a few values of "a" for which the series was zero. They were found using computers. – Mahdi Rouholamini Nov 29 '19 at 07:11
  • I would enjoy to know the webpage – Claude Leibovici Nov 29 '19 at 07:17
  • If a is not 0. For what values does it obviously not converge. – Gerben Nov 29 '19 at 07:17
  • It seems to me the series never converges. When $a\neq 0$ take Cauchy slices between $n$ and $(1+\varepsilon)n$. Then the argument in the cosine function will only vary by $a \ln(1+\varepsilon)$ so if you take $n$ big so that $\cos(a\ln(n))$ is near a multiple of $\pi$, the slice will be as big as $C\sqrt{n}$. – Carot Nov 29 '19 at 07:28
  • Here $\varepsilon$ will only depend on $a$. The case $a=0$ is obvious. The philosophy is that "$\ln$ is constant". – Carot Nov 29 '19 at 07:42
  • It diverges for $a=1$ as proved in robjohn's answer using the Euler-Maclaurin summation formula. The same method should work for most other $a$. – Conifold Nov 29 '19 at 08:08
  • This can be obviously rewritten as $Re(\zeta(1/2-ia))=0$. This is a weaker version of the Riemann hypothesis...... -_- – DinosaurEgg Nov 29 '19 at 08:12
  • No it's not. The formula for the Riemann Zeta function is only valid for $\mathfrak{R}(s)>1$. – Carot Nov 29 '19 at 08:16

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