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Let $f_{n}\left(s\right)=\frac{n!}{\left(n+1\right)^{s}}-n!$ for $n\in\mathbb{N}$ , and let $\zeta_{\mu}\left(s\right)$ denote $\sum_{m=1}^{\infty}\frac{\mu\left(m\right)}{m^{s}}$ , where $\mu\left(m\right)$ is the möbius function from number theory.

What is $\lim_{s\rightarrow0}f_{n}\left(s\right)\zeta_{\mu}\left(s\right)$ ?

This seemingly innocuous question is rife with technical subtleties.

Given a dirichlet series such as $\zeta_{\mu}\left(s\right)$, I say that I treat it analytically if I only allow $s$ to take on values for which the series is absolutely/uniformly convergent. If $s$ is to take on values at which this series does not converge, I interpret the series as representing the value obtained by performing an analytic continuation.

So, for the function $\zeta_{\mu}\left(s\right)$, treating it analytically means that $\zeta_{\mu}\left(s\right)$ can be directly evaluated at any complex $s$ with $\textrm{Re}\left(s\right)\geq1$ ; the convergence at $s=1$ (to zero, in fact) is equivalent to the Prime Number Theorem.

For $s$ not in this half-plane (ex: $s=0$ ), the analytic interpretation of the expression:

$\sum_{m=1}^{\infty}\frac{\mu\left(m\right)}{m^{0}}$

is that it denotes $\frac{1}{\zeta\left(0\right)}$ i.e., $-2$, since it is known that $\zeta_{\mu}\left(s\right)$ is equal to $\frac{1}{\zeta\left(s\right)}$ for all $s$ with $\textrm{Re}\left(s\right)\geq1$, and that the analytic continuation of $\zeta\left(s\right)$ provides for an analytic continuation of $\zeta_{\mu}\left(s\right)$.

On the other hand, I say that I treat a dirichlet series literally if I evaluate it naïvely at any s , even those at which the series is divergent. Thus, for $s=0$ , the literal interpretation of $\sum_{m=1}^{\infty}\frac{\mu\left(m\right)}{m^{s}}$ is:

$\sum_{m=1}^{\infty}\frac{\mu\left(m\right)}{m^{0}}=\sum_{m=1}^{\infty}\mu\left(m\right)$

a sum which is known to diverge to $+\infty$ (it is the limit as $n\rightarrow\infty$ of the Mertens Function $M\left(n\right)$ ).

Using an analytic interpretation of $\zeta_{\mu}\left(s\right)$ , it is easily seen that:

$\lim_{s\rightarrow\infty}f_{n}\left(s\right)\zeta_{\mu}\left(s\right)=0\times-2=0$

My question is: what is the value of this limit when $\zeta_{\mu}\left(s\right)$ is interpreted “literally”? That is, what is to be made of the expression:

$\underbrace{\left(\frac{n!}{\left(n+1\right)^{0}}-n!\right)}_{n!\times\left(1-1\right)}\underbrace{\sum_{m=1}^{\infty}\mu\left(m\right)}_{+\infty}$

Does it end up becoming 0? Or is it like a residue, and does the (literally interpreted) limit:

$\lim_{s\rightarrow0}f_{n}\left(s\right)\zeta_{\mu}\left(s\right)$ equal something non-zero?

Thanks in advance!

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    $f_n(s)$ is useless. So with $M_s(N) = \sum_{n=1}^N \mu(n)n^{-s}$ you want to know if $\lim_{N \to \infty} M_s(N)$ diverges to $\pm \infty$ or oscillates for some values of $s < 1/2$, that's it ? (it won't converge because the [abscissa of convergence](https://en.wikipedia.org/wiki/Dirichlet_series#Abscissa_of_convergence) is $\ge 1/2$) Then the case $s=0$ seems to be the one to look at : we know that $M(N) = 0$ for infinitely many $N$, so it oscillates even if $\lim \sup |M(N)|= \infty$. – 2017-02-15
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    (it oscillates because $\sum_{n=1}^N M(N/n) = 1$) – 2017-02-15
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    So is my limit divergent to ∞? Is it 0? Is it a constant? Or does it not exist? – 2017-02-15

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