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Can you prove the next theorem:

Let $f$ be Dirichlet series with real, positive coefficients $(a_n>0)$. If $f$ is holomorphic on $\Re(z)\ge1$, but has one singularity at $z=1$, then

$\lim_\limits{z \to 0^+} \left[f(z)-c(z-1)^k\right]$ exists $\iff$ $\lim_\limits{x \to \infty} \frac{A(x)\ln(x)^{k+1}}{x}=c$

where $c$ and $k$ are some real constants and

$A(x)=\sum_{n=1}^x a_n$

Behind the theorem:

I'm exploring complex analysis methods in number theory, specifically for determining asymptotic growth of partial sums of multiplicative functions. I have read Newman's proof of PNT, and I tried to use these ides to find the asymptotic growth of the next sum:

$\sum_{n=1}^{x} \frac{1}{d(n)}$

where $d(n)$ is the number of divisors of $n$. It seems to be, and computer approves it, that the growth is $\frac{cx}{\sqrt{\ln(x)}}$ for some real constant $c$. Now, in order to prove it, I defined $f$ to be Dirichlet series with coefficients $a_n=\frac{1}{d(n)}$. I proved that $f(z)-c\sqrt{\zeta(z)}$ is holomorphic for $\mathfrak{R}(z)>0$. Also, I have that $f(z)-\frac{c}{\sqrt{z-1}}$ is holomorphic for $\Re(z)\ge1$ except for $1$. It is not holomorphic at the point $1$, so I can't use Newman's analytic theorem. But, if we could ignore it, we would get that,

$\int_1^\infty \frac{A(t)-ct\sqrt{\ln(t)}}{t^2}dt$

converges. (That comes out when we differentiate $f(z)-\frac{c}{\sqrt{z-1}}$). Here I used

$A(x)=\sum_{n=1}^{x} \frac{\ln(n)}{d(n)}$.

Then, using Newman's idea, we can get that $A(x)$~$cx\sqrt{\ln(x)}$. From there, using partial summation, we get that the first sum is asymptotically equal to what it is expected to be, $\frac{cx}{\sqrt{\ln(x)}}$.

Please, help me, this is going to be in my undergraduate work, I need to know these things. Help me to prove this asymptotic growth.

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    $\begingroup$ I would suggest to take a look at korevaar's tauberian theory bible and see what's known $\endgroup$ – Conrad Mar 27 at 19:17
  • $\begingroup$ @Conrad Can you please give me a link for that? $\endgroup$ – donaastor Mar 27 at 19:31
  • $\begingroup$ This is the link to the official version of the book - any decent university library should have it and of course, I am sure that unofficial versions exist too. springer.com/us/book/9783540210580 $\endgroup$ – Conrad Mar 27 at 22:28
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I have found a work on arxiv that proves the very similar theorem and here is the link of it:

https://arxiv.org/abs/1406.0427

It is Ryo-Kato's generalization of Kable's generalization of Wiener-Ikehara Tauberian theorem. Kable states for $k$ of form $\frac{1}{n}$, and Ryo-Kato states it for all rational $k$. The real $k$ is still a mystery, though.

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  • $\begingroup$ @Conrad You actually helped me by saying that I should pay attention on Tauberian Theory. Without that I wouldn't know that I should have looked for Tauberian theorems and their extensions. So, thanks $\endgroup$ – donaastor Apr 2 at 18:34

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