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This is a question I asked myself today...

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Do you know if it is possible to build a strictly-increasing sequence $(u_n)_{n\in\mathbb{N}^\star}$ of positive integers such that $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{u_n}<+\infty$ and such that for any given $0<\varepsilon<1$, one has $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{u_n^\varepsilon}=+\infty$ ?

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This is looking for a Dirichlet series (with reciprocal of $u_n$ coefficients) with abscissa of convergence $\sigma=1$ and that has a finite limit at $s=1$, does that exist ? It would mean that the corresponding L-function (if the series was extendable around $s=1$) would have no pole at $s=1$. This is tough because the L-function would not be in the Selberg class, and those are hard to study...

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You can take $u_n = \lceil n \log^2 n\rceil$: $$ \sum_{n=1}^\infty \frac{1}{n \log^2 n} < \infty $$ but $$ \sum_{n=1}^\infty \frac{1}{n^\varepsilon \log^{2\varepsilon} n} = \infty $$ for every $\varepsilon < 1$.

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  • $\begingroup$ Oh that's totally true ! I tried $u_n=n\log(n)$ (the PNT makes sure $p_n$ works), but I didn't try that ! Thanks ! $\endgroup$
    – Anthony
    Dec 7, 2018 at 2:01
  • $\begingroup$ @Anthony Glad this helped! $\endgroup$
    – Clement C.
    Dec 7, 2018 at 2:02

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