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