Background & Question
I realised I could do a different manipulation from the one I did over here Strange method to obtain strange number theoretic identities? However after making some calculations I seems to be getting the wrong answer by perhaps an extra term.
I got the expression:
$$\implies \sum_{r=1}^n \frac{\mu(r)}{r} \ln(r) \sim \frac{\ln n}{n} -1 $$
But as $n \to \infty$ we know the $(\frac{1}{\zeta(k)})' \to 0 $ when $k$ to $1$.
Is the correct answer below?
$$\sum_{r=1}^n \frac{\mu(r)}{r} \ln(r) \sim \frac{\ln n}{n} $$
If so where did I go wrong in my proof? And what is the "normal" proof?
Proof
Let us write a relation of the Euler–Mascheroni constant for large $n$. $$ 1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots +\frac{1}{n!} \sim \gamma + \ln(n!)$$
Or, multiplying $1/2$ boths sides and let $n! \to n!/2$
$$ 0 + \frac{1}{2} + 0 + \frac{1}{4} + \dots +\frac{1}{n!} \sim \frac{\gamma}{2} + \frac{1}{2}\ln(\frac{n!}{2})$$
Or, multiplying $1/3$ boths sides and let $n! \to n!/3$
$$ 0 + 0 + \frac{1}{3} + 0 + \dots +\frac{1}{n!} \sim \frac{\gamma}{3} + \frac{1}{3}\ln(\frac{n!}{3})$$
And so on $n$ times ... Now multiplying the $r$'th row with $a_r$ and adding vertically (whilst defining $b_r$):
$$ a_1+ \frac{a_1}{2} + \frac{a_1}{3} + \frac{a_1}{4} + \dots +\frac{a_1}{n!} \sim a_1 \gamma + a_1 \ln(n!)$$
$$ 0 + \frac{a_2}{2} + 0 + \frac{a_2}{4} + \dots +\frac{a_2}{n!} \sim a_2 \frac{\gamma}{2} + \frac{a_2}{2}\ln(\frac{n!}{2})$$
$$ 0 + 0 + \frac{a_3}{3} + 0 + \dots +\frac{a_3}{n!} \sim a_3 \frac{\gamma}{3} + \frac{a_3}{3}\ln(\frac{n!}{3})$$
$$\vdots $$ $+$ $-----------------------------------$
$$ \underbrace{\frac{b_1}{1}}_{a_1/1} + \underbrace{\frac{b_2}{2}}_{(a_1+ a_2)/2} + \underbrace{\frac{b_3}{3}}_{(a_1+ a_3)/3} + \dots \sim \gamma \sum_{r=1}^n \frac{a_r}{r} + \sum_{r=1}^n \frac{a_r}{r} \ln(\frac{n!}{r}) $$
In the above we define:
$$b_r = \sum_{r|l} a_l \text{ }\forall \text{ } 1 \leq r \leq n$$ $$b_r = \sum_{(r-n)|l} a_l \text{ }\forall \text{ } n+1 \leq r \leq 2n$$ $$b_r = \sum_{(r-2n)|l} a_l \text{ }\forall \text{ } 2n+1 \leq r \leq 3n$$ $$ \vdots $$ $$b_r = \sum_{(r-(n-1)!)|l} a_l \text{ }\forall \text{ } n(n-1)!- n +1 \leq r \leq n!$$
Writing the above properly now:
$$ \sum_{r=1}^n \frac{b_r}{r} + \sum_{r=n+1}^{2n} \frac{b_{r-n}}{r} + \dots+ \sum_{r=n!-n+1}^{n!} \frac{b_{r-n!+n }}{r} \sim \gamma \sum_{r=1}^n \frac{a_r}{r} + \sum_{r=1}^n \frac{a_r}{r} \ln(\frac{n!}{r}) $$
Rearranging the L.H.S and R.H.S:
$$ \sum_{r=1}^n \sum_{k=0}^{(n-1)! -1 } b_r ( \frac{1}{kn+ r}) \sim (\gamma+ \ln n!) \sum_{r=1}^n \frac{a_r}{r} - \sum_{r=1}^n \frac{a_r}{r} \ln(r) $$
Also note : $$a_{l}=\sum _{d\mid l} \mu \left({\frac {l}{d}}\right)b_{d} \implies \frac{\partial a_l}{ \partial b_d} = \mu \left({\frac {l}{d}}\right) $$
where $\mu$ is the mobius function. Let us now act $\frac{\partial }{\partial b_d} $ on both sides where $d \leq n$:
$$ \sum_{k=0}^{(n-1)! -1 } ( \frac{1}{kn+ d}) \sim (\gamma+ \ln n!) \sum_{r=1}^n \frac{\mu(\frac{r}{d})}{r} - \sum_{r=1}^n \frac{\mu(\frac{r}{d})}{r} \ln(r) $$
where $\mu(\lambda)= 0$ where $\lambda$ is not an integer. Simplifying some of the terms:
$$ \sum_{k=0}^{(n-1)! -1 } ( \frac{1}{kn+ d}) \sim \frac{(\gamma+ \ln n!) }{n}- \sum_{r=1}^n \frac{\mu(\frac{r}{d})}{r} \ln(r) $$
Let us take $d \to 1$
$$ \sum_{k=0}^{(n-1)! -1 } ( \frac{1}{kn+ 1}) \sim \frac{(\gamma+ \ln n!) }{n}- \sum_{r=1}^n \frac{\mu(r)}{r} \ln(r) $$
Now using some approximations:
$$ 1 + \frac{1}{n} \sum_{k=1}^{(n-1)! -1 } \frac{1}{k} \sim \frac{(\gamma+ \ln n!) }{n}- \sum_{r=1}^n \frac{\mu(r)}{r} \ln(r)$$
Using the first equation with a different $n$:
$$ 1 + \frac{\gamma + \ln(n-1)!}{n} \sim \frac{(\gamma+ \ln n!) }{n}- \sum_{r=1}^n \frac{\mu(r)}{r} \ln(r)$$
$$\implies \sum_{r=1}^n \frac{\mu(r)}{r} \ln(r) \sim \frac{\ln n}{n} -1 $$