📂Functions

Proof of the Convergence of the Euler-Mascheroni Constant

Theorem

$$\gamma = \lim_{n \to \infty} \left( \sum_{k=1}^{n} \left( { 1 \over k } \right) - \ln{n} \right) = 0.577215664 \cdots$$

Description

When associated with the Riemann zeta function, it also serves as the $\gamma$ $0$’th Stieltjes constant $\gamma_{0}$. $\gamma$ is briefly known as the Euler’s constant, which has a deep relationship with the Gamma function. Setting the exact value aside, does it at least converge? Since $\ln{n}$ and the harmonic series $\displaystyle \sum_{k=1}^{n} \left( { 1 \over k } \right)$ diverges, $$\lim_{n \to \infty} \left( \sum_{k=1}^{n} \left( { 1 \over k } \right) - \ln{n} \right)$$ its existence is not evident.

Note that this number has been around for nearly 300 years, yet it’s still unknown whether it is rational or irrational.

Proof

Consider the sequence $\displaystyle \gamma _{n} := \sum_{k=1}^{n} \left( { 1 \over k } \right) - \ln{n}$.

$\gamma_{1} = 1$ and

$$\Gamma_{n} = \sum_{k=1}^{n-1} \left( { 1 \over k } \right) - \int_{1}^{n} {{1} \over {x}} dx + {{1} \over {n}}$$

Graphically, $\gamma_{n}$ is equivalent to summing up the area above $\displaystyle y = {1 \over x}$ from $x=1$ to $x=n$ and adding $\displaystyle {{1} \over {n}}$ to it.

$$\sum_{k=1}^{n-1} \left( { 1 \over k } \right) - \int_{1}^{n} {{1} \over {x}} dx > 0$$

Therefore, $\gamma _{n} > 0$. On the other hand,

$$\begin{align*} \gamma_{n+1} =& \sum_{k=1}^{n+1} {{1} \over {k}} +0 - \ln (n+1) \\ =& \sum_{k=1}^{n} {{1} \over {k}} + { 1 \over {n+1} } + \left( \ln n - \ln n \right) - \ln (n+1) \\ =& \sum_{k=1}^{n} {{1} \over {k}} - \ln n + { 1 \over {n+1} } + \ln n - \ln (n+1) \\ =& \gamma_{n} + { 1 \over {n+1} } - \ln {{n+1} \over {n}} \\ =& \gamma_{n} + { 1 \over {n+1} } - \int_{n}^{n+1} { 1 \over x } dx \end{align*}$$

since $\displaystyle { 1 \over {n+1} } < \int_{n}^{n+1} { 1 \over x } dx$,

$$\Gamma_{n+1} = \gamma_{n} + { 1 \over {n+1} } - \int_{n}^{n+1} { 1 \over x } dx < \gamma_{n}$$

That is, $\gamma_{n}$ is a decreasing sequence. As $\gamma _{n} > 0$ holds for natural numbers $n$ and $\gamma _{n}$ is a decreasing sequence, $\gamma _{n}$ converges.

■