Shorter method to derive the same result? Background
I was recently fiddling with some math and realized I could come with the below formula. However, I am unsatisfied with my proof as it is not too lengthy and I am not sure about some steps (but I am confident in the result). 
$$\gamma  = \lim_{n \to \infty}(\sum_{r=1}^{n^2}  \frac{1}{(n+r)a_{n,r}} - \ln n)$$
Where $\gamma$ is Euler's constant and $a_{n,r}= ((r−1) \mod n)+1$
Question
Can anyone else provide a shorter (another) proof for the same result? Does this converge to the the constant quicker than the original series (below)?
$$ \gamma = \lim_{n \to \infty} (1+ 1/2 + 1/3 + \dots 1/n - \ln n)$$
 A: Aside from $r=1,n+1,2n+1,3n+1,\cdots ,n(n-1)+1$ which all result $a_{n,r} =1$ the other $r$'s $1 \leq r \leq n^2$ give the result that $a_{n,r} \geq 2$.
So we want to maximize $\sum \limits_{r=1}^{n^2} \frac{1}{(n+r)a_{n,r}} $ so we know that $\sum \limits_{r=1}^{n^2} \frac{1}{(n+r)a_{n,r}} \leq \sum \limits_{r=1}^{n^2} \frac{1}{2(n+r)} + \sum \limits_{k=0}^{n-1} \frac{1}{2(n+n k+1)} = \frac{n H_{n^2+n}-H_{\frac{1}{n}}-n H_n+H_{n+\frac{1}{n}}}{2 n}$
With the fact that $ H_n \leq \ln n +1$.
We get that $\frac{n H_{n^2+n}-H_{\frac{1}{n}}-n H_n+H_{n+\frac{1}{n}}}{2 n} \leq \frac{1}{n}+\frac{\log (n)}{2 n}-\frac{\log (n)}{2}+\frac{1}{2} \log (n
   (n+1))+\frac{\log \left(n+\frac{1}{n}\right)}{2 n}+1 \leq \frac{2 \log (n)}{3}$.
So $\lim \limits_{n \to \infty} (\sum \limits_{r=1}^{n^2} \frac{1}{(n+r)a_{n,r}}-\ln n)= - \infty$
Edit : As for the second question $\frac{1}{1}+\frac{1}{2}+\cdots +\frac{1}{n}= H_n = \ln n + \gamma +\frac{1}{2 n} +O(n^{-2})$, so a better convergences is $\gamma = \lim \limits_{n \to \infty} (\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{n} - \ln n -\frac{1}{2n})$
