Existence of Euler-Mascheroni Constant

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The real sequence:

$\ds \sequence {\sum_{k \mathop = 1}^n \frac 1 k - \ln n}$

converges to a limit.

This limit is known as the Euler-Mascheroni constant.

Proof 1

Let $f: \R \setminus \set 0 \to \R: \map f x = \dfrac 1 x$.

Clearly $f$ is continuous and positive on $\hointr 1 {+\infty}$.

From Reciprocal Sequence is Strictly Decreasing, $f$ is decreasing on $\hointr 1 {+\infty}$.

Therefore the conditions of the Integral Test hold.

Thus the sequence $\sequence {\Delta_n}$ defined as:

$\ds \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$

is decreasing and bounded below by zero.

But from the definition of the natural logarithm:

$\ds \int_1^n \frac {\d x} x = \ln n$

Hence the result.


Proof 2

For $n \in \N_{>0}$ let:

$\ds \gamma_n := \sum_{k \mathop = 1}^n \frac 1 k - \ln n$


\(\ds \gamma_n\) \(=\) \(\ds 1 + \int_1^n \dfrac {\floor u} {u^2} \rd u - \ln n\) Integral Expression of Harmonic Number
\(\ds \) \(=\) \(\ds 1 + \int_1^n \dfrac {\floor u} {u^2} \rd u - \int _1 ^n \dfrac 1 u \rd u\) Definition of Real Natural Logarithm
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds 1 - \int_1^n \dfrac {u - \floor u} {u^2} \rd u\) Linear Combination of Definite Integrals
\(\ds \) \(\ge\) \(\ds 1 - \int_1^n \dfrac 1 {u^2} \rd u\) Relative Sizes of Definite Integrals as $0 \le u - \floor u < 1$
\(\ds \) \(=\) \(\ds \dfrac 1 n\)
\(\ds \) \(\ge\) \(\ds 0\)

On the other hand:

\(\ds \gamma_n - \gamma_{n + 1}\) \(=\) \(\ds \paren {1 - \int_1^n \dfrac {u - \floor u} {u^2} \rd u} - \paren {1 - \int_1^{n + 1} \dfrac {u - \floor u} {u^2} \rd u}\) by $\paren 1$
\(\ds \) \(=\) \(\ds \int_n^{n + 1} \dfrac {u - \floor u} {u^2} \rd u\) Sum of Integrals on Adjacent Intervals for Integrable Functions
\(\ds \) \(\ge\) \(\ds 0\) as $u - \floor u \ge 0$

Thus by monotone convergence theorem, the sequence $\sequence {\gamma_n}$ converges to a limit in $\R_{\ge 0}$.