Research proof index

For a continuously differentiable smoothing function supported in $[1/2,2]$ and normalized by $\int_0^\infty \nu(x)\,dx/x=1$, the Mellin factor of the smoothed cutoff at $s=1$ is $1+O(\varepsilon)$. Precisely, there is a constant $C$ such that, for every sufficiently small $\varepsilon>0$ and every $X\ge0$, $\left|\mathcal M(\widetilde{1_\varepsilon})(1)X-X\right|\le C\varepsilon X.$

Let $\gamma_3(n)=H_n-\log n-\frac{1}{2n}+\frac{1}{12n^2}-\frac{1}{120n^4},$ where $H_n$ is the $n$th harmonic number. Then $\gamma_3(n)<\gamma_3(n+1)$ for every integer $n\ge1$.

Let $\gamma_3(n)=H_n-\log n-\frac{1}{2n}+\frac{1}{12n^2}-\frac{1}{120n^4}.$ For every integer $n\ge1$, this corrected harmonic approximation is a strict lower bound for the Euler–Mascheroni constant: $\gamma_3(n)<\gamma.$