Background
I recently realised I could do a specific manipulation:
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:
$$ \sum_{r=1}^n \sum_{k=0}^{(n-1)! -1 } b_r ( \frac{1}{kn+ r}) \sim \gamma \sum_{r=1}^n \frac{a_r}{r} + \sum_{r=1}^n \frac{a_r}{r} \ln(\frac{n!}{r}) $$
Now let us take a small detour and define a number theoretic dirac delta function:
$$ \frac{\partial b_r}{ \partial a_l} = \delta( r|l)$$
Where $\delta(r|l) = 1$ when $l$ is a factor of $r$ else it is $0$. Let us now act $\frac{\partial }{\partial a_l} $ on both sides where $k \leq n$:
$$ \sum_{r=1}^n \sum_{k=0}^{(n-1)! -1 } \delta(r|l) ( \frac{1}{kn+ r}) \sim \frac{\gamma}{l} + \frac{1}{l} \ln(\frac{n!}{l}) $$
Remaining calculations
My idea was to take $n\to n+1$ in the above equation and then subtract. The R.H.S will be $ \sim \frac{1}{l}\ln (n+1)$. We can now expand both sides and get series of order $n$.
My question
Can we equate coefficients of order $n^\alpha$ where $\alpha$ is an arbitrary power? Note $n$ was originally an integer. How accurate are the resulting identities?