# Does $\lim_{n \to \infty}\sum_{k = 1}^n \zeta\Big(k - \frac{1}{n}\Big)$ equal the Euler-Mascheroni constant?

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_{11}\frac1{m^{k-\frac1n}}\right)\tag1\\&=n-1+\sum_{11}\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}

where

$$(1):$$ 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

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

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