Euler's Integral Theorem

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Theorem

$H_n = \ln n + \gamma + \map \OO {\dfrac 1 n}$

where:

$H_n$ denotes the $n$th harmonic number
$\gamma$ denotes the Euler-Mascheroni constant
$\map \OO {\dfrac 1 n}$ denotes big-$\OO$ of $\dfrac 1 n$.


Proof 1

Recall the definition of the floor function:

The floor function of $x$ is the unique integer $\floor x$ such that:

$\floor x \le x < \floor x + 1$


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

\(\ds H_n - \ln n\) \(=\) \(\ds 1 + \int_1 ^n \dfrac {\floor u} {u^2} \rd u - \ln n\) 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
\(\ds \) \(=\) \(\ds 1 - \int_1 ^n \dfrac {u - \floor u} {u^2} \rd u\) Integral Operator is Linear


Let $N \ge n \ge 1$.

By Sum of Integrals on Adjacent Intervals for Integrable Functions:

$\ds (1): \quad \paren {H_n - \ln n} - \paren {H_N - \ln N} = \int_n^N \dfrac {u - \floor u} {u^2} \rd u$

On the other hand, from Definition of Floor Function follows:

$\forall u \in \R_{\ge 1} : 0 \le \dfrac {u - \floor u} {u^2} \le \dfrac 1 {u^2}$

In view of Integral Operator is Positive, integrating the above inequality on $\closedint n N$:

\(\text {(2)}: \quad\) \(\ds 0\) \(\le\) \(\ds \int_n^N \dfrac {u - \floor u} {u^2} \rd u\)
\(\ds \) \(\le\) \(\ds \int_n^N \dfrac 1 {u^2} \rd u\)
\(\ds \) \(=\) \(\ds \dfrac 1 n - \dfrac 1 N\)
\(\ds \) \(\le\) \(\ds \dfrac 1 n\)

From $(1)$ and $(2)$, it follows:

$(3): \quad 0 \le \paren {H_n - \ln n} - \paren {H_N - \ln N} \le \dfrac 1 n$


In particular, $\sequence {H_n - \ln n}$ is a Cauchy sequence.

Thus the limit $\gamma$, the Euler-Mascheroni constant, exists by Cauchy's Convergence Criterion.


In $(3)$, for each $n \in \N$, let $N \to \infty$.

Then:

$\forall n \in \N : 0 \le H_n - \ln n - \gamma \le \dfrac 1 n$

$\blacksquare$


Proof 2

Recall the definition of the floor function:

The floor function of $x$ is the unique integer $\floor x$ such that:

$\floor x \le x < \floor x + 1$


Hence:

$0 \le x - \floor x < 1$

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

\(\ds H_n - \ln n - \gamma\) \(=\) \(\ds H_n - \ln n - \lim_{k \mathop \to +\infty} \paren {H_k - \ln k}\) Definition of Euler-Mascheroni Constant and Existence of Euler-Mascheroni Constant
\(\ds \) \(=\) \(\ds 1 + \int_1^n \dfrac {\floor u} {u^2} \rd u - \ln n - \lim_{k \mathop \to +\infty} \paren {1 + \int_1^k \dfrac {\floor u} {u^2} \rd u - \ln k}\) Integral Expression of Harmonic Number
\(\ds \) \(=\) \(\ds \int_1^n \dfrac {\floor u} {u^2} \rd u - \int_1^n \dfrac 1 u \rd u - \lim_{k \mathop \to +\infty} \paren {\int_1 ^k \dfrac {\floor u} {u^2} \rd u - \int_1^k \dfrac 1 u \rd u }\) Definition of Real Natural Logarithm
\(\ds \) \(=\) \(\ds \lim_{k \mathop \to +\infty} \paren {-\int_n^k \dfrac {\floor u} {u^2} \rd u + \int_n^k \dfrac 1 u \rd u }\) Sum of Integrals on Adjacent Intervals for Continuous Functions
\(\ds \) \(=\) \(\ds \lim_{k \mathop \to +\infty} \paren {\int_n^k \dfrac {u - \floor u} {u^2} \rd u }\) Integral Operator is Linear
\(\ds \) \(<\) \(\ds \lim_{k \mathop \to +\infty} \paren {\int_n^k \dfrac 1 {u^2} \rd u }\) Since $0 \le x - \floor x < 1$
\(\ds \) \(=\) \(\ds \intlimits {-\dfrac 1 u} n \infty\) Primitive of Power
\(\ds \) \(=\) \(\ds \paren {0 - \paren {-\dfrac 1 n} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 n\)

From Existence of Euler-Mascheroni Constant Proof 1, we have:

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

is decreasing and bounded below by zero.

Therefore:

$H_n - \ln n \ge 0$

Therefore:

$\forall n \in \N_{>0} : 0 \le \size {H_n - \ln n - \gamma} < \dfrac 1 n$

$\blacksquare$


Source of Name

This entry was named for Leonhard Paul Euler.


Sources

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