# 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.

## 1 Answer

Claim:

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.

• Thanks for your answer Jack, would you mind elaborating on how you are using equation 2? In particular how are you handling the extra logn inside the cosine in equation 1? – George Shakan Dec 28 '14 at 3:20
• @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