Euler's Integral Theorem/Proof 1
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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
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:
- $\forall n \in \N : 0 \le H_n - \ln n - \gamma \le \dfrac 1 n$
$\blacksquare$