# Landau's Theorem, Dirichlet Series

An important theorem of Landau states that, given a Dirichlet Series f(s) with coefficients $a_n$ then

If $a_n\geq0$ for all values of n, the real point of the line of convergence is a singularity of f(s).

A particular example being $\zeta(s)$, which has coefficients constantly equal to 1 and thus has a singularity on $s=1$.

But if we consider the D.S. $$f(s)=\sum_{n=1}^{\infty}\dfrac{1}{n^s\log^2{}n}$$

it clearly has non-negative Dirichlet coeffiecients $(1/\log^2{}n\geq0)$ and abscissa of convergence equal to 1 with the series $$f(1)=\sum_{n=1}^{\infty}\dfrac{1}{n\log^2{}n}$$ being convergent.

Doesn't this contrast with Landau's theorem mentioned above? What am I missing here?

• I think you want to start the sum at 2. – Chappers Jun 14 '17 at 11:06
• You are write, but still that does not answer my question. You can define $a_1=0$ and start it from 1. – Mathitis Jun 14 '17 at 11:20
• How do you prove that $f(1)$ converges? – franz lemmermeyer Jun 14 '17 at 11:27
• @franzlemmermeyer $\int \frac{dx}{x\log^2{x}} = -\frac{1}{\log{x}}$, so the integral test shows the sum converges. – Chappers Jun 14 '17 at 11:37

I suppose the meaning of singularity here is not, that $f(1)=\infty$ but rather just that f is not holomorphic at $1$. Naively differentiating $f$ leads to

$$f'(s)=-\sum_{n=2}^\infty \frac{1}{n^s\log(n)},$$

which doesn't converge at $1$ anymore.

Edit: Actually for $Re(s)>1$ the derivative definitely has the above form. Therefore $f'(1+t)\to \infty$ for $t\to 0$ so $f$ cannot be continuously differentiable at $1$, therefore it is also not holomorphic.

• This does not answer fully, because if we take the Dirichlet series $\sum_{n \ge 2} \frac{1}{n^s \log^3 n}$ the argument does not qualify. Instead, the below one is more suitable answer. – user41481 Oct 9 '17 at 7:38

Indeed, this theorem is not hard to prove.

Let $F(s) = \sum_{n=1}^\infty a_n n^{-s}, a_n \ge 0$ and $\sigma$ its abscissa of convergence, which means $\sum_{n=1}^\infty a_n n^{-x} = \infty$ for $x < \sigma$.

If $F(s)$ has an analytic continuation which is analytic around $s= \sigma$ then take $\epsilon > 0$ small enough such that the Taylor series $\sum_{k=0}^\infty \frac{F^{(k)}(\sigma+\epsilon)}{k!} (x-\sigma-\epsilon)^k$ converges for some $x < \sigma$.

With $F_N(s) = \sum_{n=1}^N a_n n^{-s}$ we obtain term-by-term $$F(x) = \sum_{k=0}^\infty \frac{F^{(k)}(\sigma+\epsilon)}{k!} (x-\sigma-\epsilon)^k \ge \sum_{k=0}^\infty \frac{F_N^{(k)}(\sigma+\epsilon)}{k!} (x-\sigma-\epsilon)^k = F_N(x)$$ a contradiction since $\lim_{N \to \infty} F_N(x) = \sum_{n=1}^\infty a_n n^{-x}=\infty$.

Qed. an analytic continuation of $F(s)$ must have a singularity at $s=\sigma$.