Let $\zeta(s)$ be the Riemann zeta function and $\gamma$ be the Euler-Mascheroni constant. I observed the following result empirically. Looking for a proof or disproof.

$$ \lim_{n \to \infty}\sum_{k = 1}^n \zeta\Big(k - \frac{1}{n}\Big) = \gamma $$

Also, I searched for different summation formulas for the Euler-Mascheroni constant using the Riemann zeta function but could not find it anywhere. Is there any reference to this sum in literature?

Update: Applying the method of @Simply Beautiful Art, we can show that

$$ \sum_{k = 1}^n \zeta\Big(k + \frac{1}{m}\Big) = \gamma + n + m + \mathcal O(n^{-1} + m^{-1}) $$


We have the simple asymptotic expansion as $s\to1$ given by:

$$\zeta(s)=\frac1{s-1}+\gamma+\mathcal O(s-1)\tag{$s\to1$}$$

For the first term of your sum, you have

$$\zeta\left(1-\frac1n\right)=-n+\gamma+\mathcal O(n^{-1})$$

and for the rest of the terms,

\begin{align}\sum_{1<k\le n}\zeta\left(k-\frac1n\right)&=\sum_{1<k\le n}\left(1+\sum_{m>1}\frac1{m^{k-\frac1n}}\right)\tag1\\&=n-1+\sum_{1<k\le n}\sum_{m>1}\frac1{m^{k-\frac1n}}\tag2\\&=n-1+\mathcal O(2^{-n})+\sum_{k>1}\sum_{m>1}\frac1{m^{k-\frac1n}}\tag3\\&=n-1+\mathcal O(2^{-n})+\sum_{m>1}\sum_{k>1}\frac1{m^{k-\frac1n}}\tag4\\&=n-1+\mathcal O(2^{-n})+\sum_{m>1}\sqrt[n]m\frac{m^{-2}}{1-m^{-1}}\tag5\\&=n-1+\mathcal O(2^{-n})+\sum_{m>1}\sqrt[n]m\left(\frac1{m-1}-\frac1m\right)\tag6\\&=n+\mathcal O(2^{-n})+\sum_{m>1}(\sqrt[n]m-1)\left(\frac1{m-1}-\frac1m\right)\tag7\\&=n+\mathcal O(n^{-1})\tag8\end{align}


$(1):$ Definition of $\zeta$.

$(2):$ Summing over $1$.

$(3):$ Extending $k$ from $(1,n]$ to $(1,\infty)$, with $\mathcal O(2^{-n})$ error.

$(4):$ Rearranging the series.

$(5):$ Geometric series.

$(6):$ Partial fractions.

$(7):$ Using telescoping series and $1=\sum_{m>1}\left(\frac1{m-1}-\frac1m\right)$.

$(8):$ Asymptotic expansion using $\sqrt[n]m=\exp(n^{-1}\ln(m))=1+\varepsilon n^{-1}\ln(m)$, where $|\varepsilon|\le\sqrt[n]m$, which gives the series bounded by $n^{-1}$ times another series with dominating term $\mathcal O(m^{\frac1n-2}\ln(m))$ and thus converges.

Adding these results together, we find that

$$\sum_{k=1}^n\zeta\left(k-\frac1n\right)=\gamma+\mathcal O(n^{-1})$$

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  • 2
    $\begingroup$ This is a nice and elegant solution, for sure ! $\endgroup$ – Claude Leibovici Aug 6 at 5:28
  • 1
    $\begingroup$ An excellent solution. Order of error term is consistent with simulation. For $n=10^9$, the sum is approximately $0.577215666086463903230516402745658$, which matches $8$ digits after decimal point. $\endgroup$ – piepie Aug 6 at 5:55
  • $\begingroup$ @piepie That looks to match the error then? $n^{-1}=10^{-9}$ so you should expect $9\pm$ a constant amount of digits accurate. $\endgroup$ – Simply Beautiful Art Aug 6 at 12:50
  • $\begingroup$ @SimplyBeautifulArt You are right. As $n$ is increased by an order of $10$, I expect 0/1/2 more digits of accuracy. $\endgroup$ – piepie Aug 7 at 3:08
  • $\begingroup$ @piepie no you should only expect 1 more digit, since $C\times10^{-(n+1)}$ is 10 times smaller than $C\times10^{-n}$. $\endgroup$ – Simply Beautiful Art Aug 7 at 13:11

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