About the Euler-Mascheroni Constant

Question

Okay, so the limiting difference between the harmonic series and the natural logarithm is known as the Euler-Mascheroni constant, $\gamma= 0,577$. My question is: is there any base for the logarithm so that the limiting difference has other results? Is it possible for the partial sums of the harmonic series minus the $\log$ in a given base of $n$ be equal to $0$ or $1$ or a number different from the Euler-Mascheroni Constant?

Answer 1

No.

Because for any base $b\ne e$, $$\ln(n)-\log_b(n)=\left(1-\frac1{\ln(b)}\right)\ln(n),$$ which diverges to $\pm\infty$.

Said differently,

$$H_n\sim a\ln(n)+b$$ requires $a=1$.

Answer 2

Take $a\in(0,\infty)\setminus\{1\}$ and, for each $n\in\mathbb N$, let $H_n$ be the $n$^th harmonic number. Then\begin{align}H_n-\log_an&=H_n-\frac{\log n}{\log a}\\&=H_n-\log n+\left(1-\frac1{\log a}\right)\log n.\end{align}But$$\lim_{n\to\infty}\left(1-\frac1{\log a}\right)\log n=\pm\infty$$unless $a=e$ and $\lim_{n\to\infty}H_n-\log n=\gamma$. Therefore your limit can be a real number only in that case.