Difference between Summation of Natural Logarithms and Summation of Harmonic Numbers

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Theorem

$\ds \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^n \map \ln {n!} \approx \gamma n + \dfrac {\ln n} 2 + 0 \cdotp 158$

where:

$H_k$ denotes the $k$th harmonic number
$n!$ denotes the $n$th factorial
$\gamma$ denotes the Euler-Mascheroni constant.


Proof

From Summation over k to n of Natural Logarithm of k:

$\ds \sum_{k \mathop = 1}^n \ln k = \map \ln {n!}$


Then:

\(\ds \map \ln {n!}\) \(\approx\) \(\ds \map \ln {\sqrt {2 \pi n} \paren {\dfrac n e}^n}\) Stirling's Formula
\(\ds \) \(=\) \(\ds \frac 1 2 \map \ln {2 \pi} + \frac 1 2 \ln n + n \paren {\ln n - \ln e}\)
\(\ds \) \(=\) \(\ds \map \ln {\sqrt {2 \pi} } + \paren {n + \frac 1 2} \ln n - n\)


Then we have that:

\(\ds \sum_{k \mathop = 1}^n H_k\) \(=\) \(\ds \paren {n + 1} H_n - n\) Sum of Sequence of Harmonic Numbers
\(\ds \) \(\approx\) \(\ds \paren {n + 1} \paren {\ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} } - n\) Approximate Size of Sum of Harmonic Series


Then:

\(\ds \) \(\) \(\ds \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^n \map \ln {n!}\)
\(\ds \) \(\approx\) \(\ds \paren {\paren {n + 1} \paren {\ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} } - n} - \paren {\ln \paren {\sqrt {2 \pi} } + \paren {n + \frac 1 2} \ln n - n}\)
\(\ds \) \(=\) \(\ds n \ln n + n \gamma + \dfrac 1 2 - \dfrac 1 {12 n} + \dfrac 1 {120 n^3} + \ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - n - \map \ln {\sqrt {2 \pi} } - n \ln n - \frac 1 2 \ln n + n\)
\(\ds \) \(=\) \(\ds \gamma n + \frac 1 2 \ln n + \dfrac 1 2 + \gamma - \map \ln {\sqrt {2 \pi} } + \dfrac 5 {12 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^3} + \dfrac 1 {120 n^4}\)


The result follows by ignoring the lower order terms in $n$.

$\blacksquare$


Sources

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