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.
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