Question on Macys formula for Euler-Mascheroni Constant $\gamma$

Question

I think that:

$\gamma = \lim_{n\rightarrow\infty} ~~~ 2H_{n} - H_{n(n+1)}~~~~~~$ (where $H_{n}$ is the $n$-th harmonic number)

is a closed form of Macys $\gamma$ formula:

$\gamma = \lim_{n\rightarrow\infty} ~~~ (1+\frac{1}{2}+...+\frac{1}{n}-\frac{1}{n+1}-...-\frac{1}{n^{2}}-\frac{1}{n^{2}+1}-...-\frac{1}{n^{2}+n})$

which I stumbled upon in Will Jagy's comment to the M.SE question:

What is the fastest/most efficient algorithm for estimating Euler's Constant?

I tried to read the references given to Macys paper, but they appear to be all behind a paywall. My question is, if this closed form is already mentioned in those papers?

Answer 1

$$\gamma_n=H_n-\ln n=H_n-k\frac{\ln n}k=H_n-\frac{\ln n^k}k=H_n-\frac{H_{n^k}-\gamma_{n^k}}k\iff$$ $$\iff\lim_{n\to\infty}\left(H_n-\dfrac{H_{n^k}}k\right)=\dfrac{k-1}k\cdot\gamma$$