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

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?

• Can you give the bibliographic details for Macy's paper? – Gerry Myerson Dec 13 '13 at 11:17
• Bib-reference should be this: J. J. Mačys, “On the Euler–Mascheroni Constant”, Mat. Zametki, 94:5 (2013), 695–701 – nyc Dec 13 '13 at 11:32
• By the way, it also seems that $\gamma ~ =~\lim_{n\rightarrow\infty}~\frac{kH_{n}-H_{n^{k}}}{k-1}$ for all $k\in\mathbb{N}\geq2$ – nyc Dec 13 '13 at 11:34
• Have you checked to see whether the author posted a version on arXiv? – Gerry Myerson Dec 13 '13 at 11:35
• Did not yet found it on arXiv. – nyc Dec 13 '13 at 11:37

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

• Thanks for the proof. I will wait for a couple of days, if anybody adresses the original question and if not, accept yours as answer. – nyc Dec 14 '13 at 7:51
• @nyc: Here is the first page of a Russian version of the article in question, containing the formula. Knowing the meaning of Cyrillic letters, I can confirm that the name there is indeed Ma$\breve{\text{c}}$is. – Lucian Dec 14 '13 at 21:32
• Thanks. The closed form formula isn't mentioned on the first page. Could you maybe read any statement in the text on that page, if Macis mentions a closed form? – nyc Dec 15 '13 at 5:26
• @nyc: Unfortunately I don't speak any Russian. :-) I only know the meaning of their letters. But this guy might be able to help you... Actually, I just used Google Translate, and wasn't able to find the expression "closed form" there. – Lucian Dec 15 '13 at 8:44

Here is a proof for your expression.

Given $$\gamma=\lim_{n \to \infty}\left(H_n-log(n)\right)$$

and $$\lim_{n \to \infty} log\left( \frac{n^2}{n(n+1)}\right)=0$$

we have $$lim_{n \to \infty}\left(2H_n-H_{n(n+1)}\right) = lim_{n \to \infty}\left(2H_n-H_{n(n+1)}\right)- \lim_{n \to \infty} log\left( \frac{n^2}{n(n+1)}\right)$$ $$=lim_{n \to \infty}\left(2H_n-H_{n(n+1)}- log\left( \frac{n^2}{n(n+1)}\right)\right)$$ $$=lim_{n \to \infty}\left(2H_n-2log\left(n\right)-H_{n(n+1)}+log\left(n(n+1)\right)\right)$$ $$=2\gamma-\gamma=\gamma$$