Dirichlet Series and Euler Products -- Examples

For the zeta series, fix $s$ with $\mathrm{\Re }s>1$ and choose a rational $q$ with $1<q<\mathrm{\Re }s$. Then $\mathrm{\mid }{n}^{-s}\mathrm{\mid }=n^{-\mathrm{\Re }s}\le n^{-q}$, so [L5] gives absolute convergence. At $s=1$ the same series is the harmonic series and diverges by [L5]. Therefore [L1], [L2], and [L3] force both abscissae to equal $1$: convergence at any point with real part $<1$ would imply convergence at $1$, and absolute convergence at any point with real part $<1$ would imply absolute convergence at $1$.

For the eta series, fix $s$ with $\sigma :=\mathrm{\Re }s>0$. The partial sums of $\sum (-1{)}^{n-1}$ are bounded by $1$. Applying [L4] to the tail weights $b_n=n^{-s}$ gives ${\sum }_{n=M}^N(-1{)}^{n-1}{n}^{-s}=A_N{N}^{-s}-A_{M-1}{M}^{-s}+{\sum }_{n=M}^{N-1}{A}_n(n^{-s}-(n+1{)}^{-s}),$ with $\mathrm{\mid }{A}_n\mathrm{\mid }\le 1$. Since $n^{-s}-(n+1{)}^{-s}=O_s(n^{-\sigma -1})$ and $N^{-s}\to 0$, the right-hand side tends to $0$ as $M,N\to \mathrm{\infty }$, so the eta series converges for every $\mathrm{\Re }s>0$. Its absolute series is $\sum n^{-\mathrm{\Re }s}$, so step 1.1 shows ${\sigma }_a=1$. At $s=0$ the terms are $(-1{)}^{n-1}$, which do not tend to $0$, so [L6] gives divergence. Therefore [L1] and [L2] force ${\sigma }_c=0$.

The first Möbius values are $\mu (1)=1,\mu (2)=\mu (3)=\mu (5)=\mu (7)=-1,\mu (4)=\mu (8)=0,\mu (6)=1.$ So the displayed initial segment is exactly ${\sum }_{n\le 7}\mu (n)n^{-s}$.

Multiplying this initial segment by the initial segment of $\zeta (s)$ and using [L2], the coefficients through $n=7$ cancel to those of the Dirichlet-convolution identity. That is the coefficient-level content of [L1].

The positive divisors of $12$ are $1,2,3,4,6,12$, so $\tau (12)=6$.

Therefore [L1] says that the coefficient of ${12}^{-s}$ in $\zeta (s{)}^2$ is $6$.

By [L2], ${\sum }_{k\ge 0}\phi (p^k)p^{-ks}=1+{\sum }_{k\ge 1}(p^k-p^{k-1})p^{-ks}=1+(1-p^{-1}){\sum }_{k\ge 1}{p}^{k(1-s)}\mathrm{.}$

Summing the geometric series gives $1+(1-p^{-1})\frac{p^{1-s}}{1-p^{1-s}}=\frac{1-p^{-s}}{1-p^{1-s}}\mathrm{.}$ This is the local factor of [L1].

By [L2], ${\sum }_{n\ge 1}\lambda (n)n^{-s}={\prod }_p\frac{1}{1+p^{-s}}\mathrm{.}$ Indeed [L1] gives the local ratio $f(p)=-1$, so the denominator is $1-(-p^{-s})=1+p^{-s}$.

Since $\frac{1}{1+p^{-s}}=\frac{1-p^{-s}}{1-p^{-2s}},$ multiplying over primes gives ${\prod }_p\frac{1-p^{-s}}{1-p^{-2s}}=\frac{\zeta (2s)}{\zeta (s)}\mathrm{.}$ So the claimed identity holds.

On $\mathrm{\Re }s>1$, [L1] gives $\zeta (s)={\prod }_p(1-p^{-s}{)}^{-1}\mathrm{.}$ At the boundary point $s=1$, however, the Dirichlet series is the harmonic series and diverges. So there is no value of the left-hand side there to which the proved Euler-product identity could apply.

This already refutes the claimed boundary extension: the theorem proving the Euler product does not license prime-factor regrouping once absolute convergence is lost.

If $\mathrm{\Re }s=\sigma >0$, then ${\sum }_{k\ge 1}\mid \frac{2^{-ks}}{k}\mid \le {\sum }_{k\ge 1}{2}^{-k\sigma },$ and the right-hand side is a convergent geometric series by [L3]. So $D(s)$ converges for every $\mathrm{\Re }s>0$. At $s=0$ it becomes the harmonic series ${\sum }_{k\ge 1}1\mathrm{/}k$, which diverges by [L2]. Therefore [L1] gives ${\sigma }_c=0$.

On the same boundary line, at $s_{\ast }=\frac{\pi i}{\log 2}$ one has $2^{-s_{\ast }}=-1$, so $D(s_{\ast })={\sum }_{k\ge 1}\frac{(-1{)}^k}{k},$ which converges by [L2]. Thus the line $\mathrm{\Re }s=0$ contains both the divergent point $s=0$ and the convergent point $s=s_{\ast }$.

Therefore convergence at one boundary point does not force convergence at every point of the abscissa line.