# Abscissa of convergence for a Dirichlet series

Let $\alpha \in \mathbb{Z}$ and $f(n) = n^{i \alpha n}$. What is the abscissa of convergence, $\sigma_c$, for the associated Dirichlet series, $\sum_{n=1}^{\infty} \frac{f(n)}{n^s}$? Since $|f(n)| = 1$, it follows that $\sigma_a = 1$ (where $\sigma_a$ is the abscissa of absolute convergence), and we may conclude from general theory of Dirichlet series that $\sigma_c \in [0,1]$.

My feeling is that there should be a bit of cancellation, resulting in $\sigma_c$ being smaller than 1, though I haven't been able to quantify this.

Given that $\alpha\in\mathbb{Z}\setminus\{0\}$, $$\sum_{n=1}^{+\infty}\frac{n^{i\alpha n}}{n^s}$$ is convergent for any $s$ such that $\Re(s)>0$.
Proof: It is sufficient to prove that both $$C_N = \sum_{n=1}^{N}\cos(\alpha n\log n)\quad\text{and}\quad S_N=\sum_{n=1}^{N}\sin(\alpha n\log n)$$ are $\ll\log^2(N)$, then apply partial summation. We have: $$C_N = \sum_{k=0}^{\lfloor\log N\rfloor}\sum_{n\in(e^k,e^{k+1})}\cos(\alpha n\log n) = \sum_{k=0}^{\lfloor\log N\rfloor}A_k\tag{1}.$$ Since for any $\beta\in\mathbb{Z}^+$ we have: $$\max\left(\left|\sum_{n=M}^{M+N}\sin(\beta n)\right|,\left|\sum_{n=M}^{M+N}\cos(\beta n)\right|\right)\leq\frac{1}{\sin(\beta/2)}\leq\left\|\frac{\beta}{2\pi}\right\|^{-1}\tag{2}$$ where $\|x\|$ is the distance between $x$ and the closest integer, it is not difficult to prove the logarithmic bound for $C_N$ and $S_N$. In a somewhat less quantitative way, by Van Der Corput's difference theorem we know that the sequence $\{\alpha n\log n\}_{n\geq 1}$ is almost equidistributed $\pmod{2\pi}$: by approximating its discrepancy is it possible to recover an upper bound for $C_N$ and $S_N$ in terms of some power of $\log N$, that is enough to let us state that the abscissa of conditional convergence is zero.
• @GeorgeShakan: I am approximating $\log n$ for $n\in(e^k,e^{k+1})$ with $k$, simply. Then we have to bound the sum of $\cos((\alpha k) n)$ over this interval. – Jack D'Aurizio Dec 28 '14 at 3:39
• What is your error term when you approximate $cos(nαlogn)$ by $cos(nαk)$? I think it is unreasonable to expect anything better than square root cancellation in this type of problem. – George Shakan May 23 '15 at 5:47