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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 $\ln(n)$?

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    $\begingroup$ Good question, I never thought about that... $\endgroup$ – Troy Woo Aug 27 '14 at 14:28
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    $\begingroup$ Yes, they are identical (in the limit). Compute $\lim\limits_{n\to\infty} \log(n+1)-\log(n)$. $\endgroup$ – Ian Mateus Aug 27 '14 at 14:28
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    $\begingroup$ It makes it possible (?) to write $\displaystyle\left(\sum_{k=1}^n-\int_1^n dk\right)\frac 1 k$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 27 '14 at 14:50
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    $\begingroup$ In fact subtracting $\;\ln\left(n+\frac 12\right)\,$ is the best choice (the error made is near $\dfrac 1{2n}$ for $-\ln(n+0)$ and $-\ln(n+1)$ while near $\dfrac 1{24\,n^2}$ for $+\dfrac 12$). $\endgroup$ – Raymond Manzoni Aug 27 '14 at 14:51
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    $\begingroup$ @MGA: I considered $\delta(n,a):= \sum_{k=1}^n \frac{1}{k} - \ln(n+a)-\gamma\;$ then $\,\delta(10,0) \approx 0.049167496$ (near $\frac 1{20}$) while $\delta(10,1) \approx -0.046142683$ (near $-\frac 1{20}$) and $\delta(10,1/2) \approx 0.00037733190$ (near $\frac 1{2650}$). Hoping this clarified things, $\endgroup$ – Raymond Manzoni Aug 27 '14 at 18:18
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$$ \left( \sum_{k=1}^{n} \dfrac{1}{k} - \ln(n+1) \right) - \left( \sum_{k=1}^{n} \dfrac{1}{k} - \ln(n) \right)=\ln(n)-\ln(n+1)=\ln\left(\frac{n}{n+1}\right)$$

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

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  • $\begingroup$ Thank you, good answer. But there is still something strange for me: In the above, we could use $\ln(n+A)$ for any finite $A\in \mathbb{N}$, and the limit would still evaluate to $0$, but it's certainly not true in general that $\sum_{k=1}^{n} 1/k > \ln(n+A)$. I wonder what I'm missing? $\endgroup$ – MGA Aug 27 '14 at 14:33
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    $\begingroup$ @MGA While it is not true in general that we have $\sum_{k=1}^{n} \dfrac{1}{k} > \ln(n+A)$ for all $n$, we always have $\sum_{k=1}^{n} \dfrac{1}{k} > \ln(n+A)$ for $n$ large enough. $\endgroup$ – N. S. Aug 27 '14 at 14:37
  • $\begingroup$ Thanks again, I'll try to wrap my head around that. I'm new to analysis, so I'm not use to being very rigorous. $\endgroup$ – MGA Aug 27 '14 at 14:39
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    $\begingroup$ @MGA To make it simple, it is the same phenomena which happens with lets say $1-\frac{1000000}{n}$. This sequence converges to $1$, even if the first few terms are negative. It eventually becomes positive..... In general, if a sequence converges to a strictly positive number, it means that from some step further all terms are positive (but it can be negative for a while). A finite number of terms doesn't matter for convergence. $\endgroup$ – N. S. Aug 27 '14 at 14:41
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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.

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