Let $\zeta(s)$ be the Riemann zeta function and $\gamma$ be the Euler-Mascheroni constant. Is the following formula for the Euler-Mascheroni constant true?
$$ \lim_{n \to \infty}\sum_{k=1}^n \left[\zeta\left(2k-1-\frac{1}{2n}\right) + \zeta(2k)\right] = \gamma $$
Yes, and the method is the same. Writing $f(n)\approx g(n)$ for $f(n)=g(n)+\mathcal{O}(n^{-1})$ as $n\to\infty$, \begin{align*} \sum_{k=1}^n\left[\zeta\left(2k-1-\frac{1}{2n}\right)+\zeta(2k)\right] &\approx-2n+\gamma+\zeta(2)+\sum_{k=2}^n\sum_{m\geqslant 1}(m^{1+1/(2n)}+1)m^{-2k} \\&=-2+\gamma+\zeta(2)+\sum_{k=2}^n\sum_{m>1}(m^{1+1/(2n)}+1)m^{-2k} \\&\approx-2+\gamma+\zeta(2)+\sum_{m,k>1}(m^{1+1/(2n)}+1)m^{-2k} \\&=-2+\gamma+\zeta(2)+\sum_{m>1}\frac{m^{1+1/(2n)}+1}{m^2(m^2-1)} \\&=-2+\gamma+\zeta(2)+\sum_{m>1}\frac{1}{m^2(m-1)}+\sum_{m>1}\frac{m^{1/(2n)}-1}{m(m^2-1)} \\&\approx-2+\gamma+\zeta(2)+\sum_{m>1}\frac{1}{m^2(m-1)}, \end{align*} and the last sum is equal to $$\sum_{m>1}\left(\frac{1}{m-1}-\frac{1}{m}\right)-\sum_{m>1}\frac{1}{m^2}=1-\big(\zeta(2)-1\big)=2-\zeta(2).$$