Is $\lim_{s \rightarrow 1} E(f(X_s)) = \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k=1}^N f(k)$? Let $s>1$ be a real number. We look at the zeta probability function / Zipf probability function defined as:
$$P(X = n) = \frac{1}{n^s \zeta(s)}$$
Suppose $f: \mathbb{N} \rightarrow \mathbb{R}$ is a function such that the following limit exists:
$$\lim_{s \rightarrow 1} E(f(X_s)) = \lim_{s \rightarrow 1} \lim_{N\rightarrow \infty} \sum_{k=1}^N f(k) P(X_s =k ) =$$
$$ = \lim_{s \rightarrow 1} \lim_{N\rightarrow \infty} \sum_{k=1}^N f(k) \frac{1}{k^s \zeta(s)} = \lim_{s \rightarrow 1} \frac{1}{\zeta(s)}\lim_{N\rightarrow \infty} \sum_{k=1}^N \frac{f(k)}{k^s} = \lim_{s \rightarrow 1} \frac{D_f(s)}{\zeta(s)}$$
where $D_f(s) = \sum_{k=1}^\infty \frac{f(k)}{k^s}$ is a Dirichlet series which converges for $s>1$ and the limit $= \lim_{s \rightarrow 1} \frac{D_f(s)}{\zeta(s)}$ is defined and exists.
Question: Is $\lim_{s \rightarrow 1} E(f(X_s)) = \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k=1}^N f(k)$?
If this can be proven, then the Prime number theorem would follow by plugging in $f(X_s) = \lambda(X_s)$ where $\lambda(n) = (-1)^{\Omega(n)}$ is the Liouville function, since it is known since Landau's doctoral thesis in 1899 that the right hand side
$$\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k=1}^N \lambda(k) = 0$$
is equivalent to the Prime number theorem.
But if the above question can be positively answered, then, since by Wikipedia we know that the left hand side is equal to:
$$\lim_{s \rightarrow 1 }\frac{D_{\lambda}(s)}{\zeta(s)} = \lim_{s \rightarrow 1} \frac{\zeta(2s)}{\zeta(s)^2} = \frac{\zeta(2)}{\infty} = 0$$
so the right hand side would also be equal to $0$ and the prime number theorem would follow.
Intuitively for $s \rightarrow 1$ the probability
$$\lim_{s \rightarrow 1} P(X_s \equiv 0 \mod(n)) = \lim_{s \rightarrow 1} \frac{1}{n^s} = \frac{1}{n}$$
equals the "probability that a random uniform natural number $X$ is divisible by $n$", which is $1/n$. So that is why I would expect the question to be positively answered, since it is a question about two expected values:
$$\lim_{s \rightarrow 1} E(f(X_s)) =^? \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k=1}^N f(k) = \lim_{N \rightarrow \infty} E(f(X_N))$$
where $X_N$ is a random number $1 \le X_N \le N$ drawn uniformly with probability $1/N$.
 A: The answer is no. E.g., if
$$f(k)=\sum_{j=0}^\infty 1(2^{2j}<k\le2^{2j+1})$$
for natural $k$, then
\begin{equation}
    Ef(X_s)\to\frac13 \tag{1}\label{1}
\end{equation}
as $s\downarrow0$, whereas
$$\frac1n\,\sum_{k=1}^n f(k)$$
will be forever oscillating between $\frac13$ and $\frac23$ as $n\to\infty$.
Indeed, let $s\downarrow0$. Then $\zeta(s)\sim\frac1{s-1}$ and
\begin{equation}
\begin{aligned}
    \zeta(s)Ef(X_s)&=\sum_{k=1}^\infty\frac{f(k)}{k^s} \\  
    &=\sum_{k=1}^\infty\frac1{k^s}\sum_{j=0}^\infty 1(2^{2j}<k\le2^{2j+1}) \\  
    &=\sum_{j=0}^\infty\sum_{k=1}^\infty\frac1{k^s} 1(2^{2j}<k\le2^{2j+1}) \\  
    &=\sum_{j=0}^\infty\Big(\int_{2^{2j}}^{2^{2j+1}}\frac{dk}{k^s}
    +O\Big(\frac1{2^{2sj}}\Big) \Big) \\  
    &=\sum_{j=0}^\infty\frac{1-2^{1-s}}{s-1}\,2^{(1-s)2j}+O(1) \\  
    &\sim\frac1{3(s-1)}, 
\end{aligned}
\end{equation}
so that \eqref{1} is proved.
On the other hand, letting
\begin{equation}
    s(n):=\sum_{k=1}^n f(k) 
\end{equation}
and letting a natural $m$ go to $\infty$, we have
\begin{equation}
\begin{aligned}
    s(2^{2m+2})&=s(2^{2m+1}) \\  
    &=\sum_{k=1}^{2^{2m+1}}\sum_{j=0}^\infty 1(2^{2j}<k\le2^{2j+1}) \\  
    &=\sum_{k=1}^{2^{2m+1}}\sum_{j=0}^m 1(2^{2j}<k\le2^{2j+1}) \\  
    &=\sum_{j=0}^m\sum_{k=1}^{2^{2m+1}} 1(2^{2j}<k\le2^{2j+1}) \\  
    &=\sum_{j=0}^m 2^{2j}\sim\frac43\,2^{2m},   
\end{aligned}
\end{equation}
so that
\begin{equation}
    \frac{s(2^{2m+2})}{2^{2m+2}}\to\frac13,
\end{equation}
whereas
\begin{equation}
    \frac{s(2^{2m+1})}{2^{2m+1}}\to\frac23.
\end{equation}
So,
\begin{equation}
    \frac1n\,\sum_{k=1}^n f(k)=\frac{s_n}n
\end{equation}
oscillates (at least) between $\frac13$ and $\frac23$ as $n\to\infty$, as was  claimed.
