Let $D(s) = \sum\_{n=1}^\infty a\_n n^{-s}$ be a Dirichlet series with $a\_n ≥ 0$ and abscissa of convergence $\sigma\_a = 1$. Further, we assume that $D(s)$ is holomorphic in each point $\Re(s) = 1$ except a pole of order $k > 0$ in $s = 1$ and that $D(s)$ possesses a meromorphic continuation on some half-plane $\Re(s) > 1 - \epsilon\_0$ with some $\epsilon\_0 > 0$. 


My question is, whether under the above assumptions we can already say that


$$ \sum\_{n ≤ x} a\_n = xP(\log x) + o(x)$$


with some polynomial $P$ with degree $k-1$. Unfortunately I am not able to exclude other poles in the strip $1 > \Re(s) > 1 - \epsilon\_0$.


In the simple case $k=1$ the question follows by the Wiener-Ikehara theorem, since then we have $\sum\_{n ≤ x} a\_n \sim Cx$ with $C = \mathrm{res}\_{s=1}D(s)$.


Thank you!



A question concerning Tauberian theory
The answer is no, even if we assume there are no other poles than $1$ in $\sigma > 1- \epsilon\_0$. I give an example below with $\epsilon\_0=1$. This is a variant of an example given by Karamata in $1952$.


Let $b\_n = 1 + \cos(\log^2(n)) \geq 0$ and consider $B(s) = \sum\_{n \geq 1} b\_n n^{-s}$. Let us write
$$
\sum\_{n \leq x} b\_n = \int\_{1}^x (1+\cos(\log^2(t))) dt + R(x)
$$
where $R(x) = O(\log^2(x))$. Then for $\sigma >1$ we have
$$
B(s) = \int\_{1}^{+ \infty}(1+ \cos(\log^2(t)) )t^{-s}dt + s \int\_{1}^{+ \infty} R(t) t^{-s-1} dt.
$$
The last integral extends holomorphically to $\sigma >0$. The first integral is equal to 
$$
\frac{1}{s-1} + \frac{1}{2} I\_+(s) + \frac{1}{2} I\_-(s)
$$
where 
$$
I\_{\pm}(s) = \int\_{0}^{+ \infty} \exp{((1-s)u \pm i u^2)} dt.
$$
Using contour integration one can move the integration line to $u \mapsto e^{\pm \frac{i \pi}{4}}u$, and this yields 
$$
I\_{\pm}(s) = e^{\pm \frac{i \pi}{4}} \int\_{0}^{+ \infty} \exp{((1-s)u e^{\pm \frac{i \pi}{4}} - u^2)} dt.
$$
Thus $I\_+$ and $I\_-$ are entire functions.


Let us now choose $a\_n = b\_n \log(n) \geq 0$ so that $D(s) = - B'(s)$ is meromorphic on $\sigma >0$, with a unique pole at $s=1$ of order $2$. One checks that
$$
\sum\_{n \leq x} a\_n = x \log(x) + x \left( \frac{1}{2} \sin(\log^2(x))-1 \right) +O(\frac{x}{\log x})
$$
**Remark** : The error term in Delange's theorem can be improved provided growth assumptions (on a strip) are made on $D(s)$.


