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.
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.