Why is there a 'missing' $1$ in the Euler–Mascheroni constant?

Question

It is easy to show that: $$ \sum_{k=1}^n \frac{1}{k} > \ln(n+1), $$ but the Euler–Mascheroni constant is defined as: $$ \gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \ln(n) \right). $$

My question is, why was $\gamma$ defined using $\ln(n)$ and not $\ln(n+1$)?

Are the two definitions identical, or does it simply turn out to be more convenient for other applications to define $\gamma$ using $\ln(n)$?

Answer 1

$$ \left( \sum_{k=1}^{n} \dfrac{1}{k} - \ln(n+1) \right) - \left( \sum_{k=1}^{n} \dfrac{1}{k} - \ln(n) \right)=\ln(n)-\ln(n+1)=\ln\left(\frac{n}{n+1}\right)$$

And $$\lim_n \ln\left(\frac{n}{n+1}\right)=\ln 1=0$$

Answer 2

As @N.S. pointed out, it does not make any difference in the limit by choosing $\ln(n+A)$ rather than $\ln(n)$.

The use of $\ln(n)$ comes from the convenient integral formula given by $$\gamma = \int_1^\infty \left( \frac{1}{[x]} - \frac1x \right) dx.$$

Here the choice to use $\ln(n)$ is natural.

Moreover, we can view $\gamma$ as the difference in the areas of $1/x$ and $1/[x]$. It can be seen from this viewpoint, by sliding all of the areas under $1/[x]$ and over $1/x$ to the $y$-axis, that $\gamma$ is bounded by $1$. Since all of these areas fit inside of the square that has the origin as well as $(1,1)$ for corners.