Why does Dirichlet Series of Mangoldt Function has simple pole of order 1 at s = 1

Question

Could someone explain why $\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s} = -\frac{\zeta'(s)}{\zeta(s)}$ has a first order pole at $s=1$ with residue 1? That's what I found from Apostol's Introduction to Analytic Number Theory, Chapter 13.1. Thank you.

Answer 1

Presuming that you know that $\zeta(s)$ has a pole of order 1 with residue 1 at 1, then we know that $(s-1)\zeta(s)\rightarrow1$ as $s\rightarrow1$, and the Laurent series of $\zeta$ at 1 has principal part $\frac{1}{s-1}$. Thus the Laurent series of $-\zeta^{\prime}$ has principal part $\frac{1}{(s-1)^2}$. So $-\zeta^{\prime}$ has a pole of order 2 with residue 1 at 1. Therefore the quotient satisfies, $(s-1)\frac{-\zeta^{\prime}(s)}{\zeta(s)}=\frac{(s-1)^2(-\zeta^{\prime})(s)}{(s-1)\zeta(s)}\rightarrow\frac{1}{1}=1$ as $s\rightarrow1$. So $-\frac{\zeta^{\prime}}{\zeta}$ has a pole of order 1 with residue 1.

Answer 2

As discussed HERE, the Riemann Zeta function $\zeta(s)$ has a pole of order $1$ at $s=1$. Since it is analytic in the complex plane except for $s=1$, we can write

$$\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^\infty a_n(s-1)^n \tag 1$$

Differentiating $(1)$, we find

$$\zeta'(s)=-\frac{1}{(s-1)^2}+\sum_{n=1}^\infty na_n(s-1)^{n-1} \tag 2$$

Using $(1)$ and $(2)$ together reveals

$$-\frac{\zeta'(s)}{\zeta(s)}=\frac{\frac{1}{s-1}-\sum_{n=1}^\infty na_n(s-1)^n}{1+\sum_{n=0}^\infty a_n(s-1)^{n+1}} \tag 3$$

To see that the ratio in $(3)$ has a pole of order $1$ at $s=1$ with residue $1$ we use the right-hand side of $(3)$ to evaluate the limit $\lim_{s\to 1}(s-1)\left(-\frac{\zeta'(s)}{\zeta(s)}\right)$. Evidently, we have

$$\lim_{s\to 1}(s-1)\left(-\frac{\zeta'(s)}{\zeta(s)}\right)=1$$

And we are done!

Answer 3

You can use the following generalized result:

Let $f$ be a meromorphic function on an open set $\Omega$ that does not vanish identically. Then the only poles of $f'/f$ are simple poles (poles of order $1$), occurring at the poles and zeros of $f$ (after all removable singularities have been removed). Furthermore, the residue of $f'/f$ at a pole $z_0$ is an integer, equal to the order of zeo of $f$ if $f$ has a zero at $z_0$, or equal to negative the order of a pole at $z_0$ if $f$ has a pole at $z_0$.

The proof is quite easy, just write $f$ as $f(z)=(z-z_0)^mF(z)$ with $F(z_0)\neq 0$ if $f$ has a zero at $z_0$ or $f(z)=F(z)/(z-z_0)^m$ with $F(z_0)\neq 0$ if $f$ has a pole at $z_0$, then computing the log derivative $f'/f$.

Since zeta function $\zeta$ has a pole of order $1$ at $s=1$, by the previous result, we see that $\zeta'/\zeta$ has a simple pole at $s=1$ with residue $-1$.