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Let $$ C_n=\sum_{i=1}^n \frac{1}{i}-\log n,\quad n\geq 1 $$ It is well-known that the limit of $(C_n)_{n\geq 1}$ exists, where its limit is denoted $\gamma$, Euler's constant. I'll shortly write what the author Artin has stated this proof in his work, Gamma function. He defines $D_n=C_n-\frac{1}{n}$ for $n\geq 1$. Then, he proceeds proving that $(C_n)_{n\geq 1}$ is strictly increasing and $(D_n)_{n\geq 1}$ is strictly decreasing. Since $D_n<C_n$, it follows that $(C_n)_{n\geq 1}$ is bounded below by $D_1=0$, and thus the sequence converges. That's all.

The question is:

How to show that $\gamma$ lies in $(0,1)$?

Since $0=D_1<D_n<C_n<C_1=1$, and since limits do not preserve the strict inequalities, the limit must lie in $[0,1]$. Is there a (hopefully simpler) way to transform from $[0,1]$ to $(0,1)$?

Mr.MathDoctor
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1 Answers1

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A graphical proof (with its advantages and drawbacks : "is it really a proof ? etc.) needing only to know that $\ln(k)$ is the area under the curve of $y=\dfrac{1}{x}$ taken from $x=1$ to $x=k$ :

enter image description here

Indeed, if we set

$$E_n:=C_n-1=\sum_{i=2}^n \dfrac{1}{i}$$

$E_n$ can be visualized as the sum of the areas of the blue curvilinear right triangles from $1$ to $n$ taken as negative areas. You can slide them (without overlap) in order they fit into the square $[0,1] \times [0,1]$. As they are considered negatively, it is the complement set, the white part, that represents the sequence ; this answers the question because the limit set, a kind of fractal set, has a limit area slightly larger than $0.5$... ($\gamma \approx 0.577...$) due to the concavity of the curvilinear triangles.

Jean Marie
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  • Good. Can you shed any light on this idea used in France? https://math.stackexchange.com/questions/4391704/what-are-suites-adjacentes-called-in-english It is my favorite way to estimate Euler's $\gamma,$ as $a_n = H_n - \log n $ and $b_n = H_{n+1} - \log n $ because these are "adjacent sequences," for any positive integers $m,n,$ we have $a_n < \gamma < b_n $ – Will Jagy Mar 01 '22 at 02:39
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    @Will Jagy It is indeed a "specialized" term that was used in french mathematics curriculum. it would be interesting to track its origin. I wouldn't be surprized it is a term coined at the time of Cauchy. – Jean Marie Mar 01 '22 at 08:02