Theorem: Approximate Size of Sum of Harmonic Series

Let $H_n$ denote the sum of the harmonic series:
:$H_n = \ds \sum_{k \mathop = 1}^n \frac 1 k$
Then $H_n$ can be approximated as follows:
:$H_n \approx \ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - \epsilon$
where:
:$\gamma$ denotes the Euler-Mascheroni constant: $\gamma \approx 0 \cdotp 57721 \, 56649 \, \ldots$
:$0 < \epsilon < \dfrac 1 {252 n^6}$

Proof:
{{tidy}}
{{begin-eqn}}
{{eqn | l = H_n
      | r = \sum_{i=1}^n\frac{1}{k}
      | c = Definition of harmonic numbers
}}
{{eqn | r = \int_1^n \frac 1 x \rd x + \sum_{k \mathop = 1}^\infty \frac {B_k} {k!} \paren {\map {f^{\paren {k - 1} } } n - \paren {-1}^k \map {f^{\paren {k - 1} } } 1}
      | c = Euler-Maclaurin Summation Formula
}}
{{eqn | r =  \map \ln n + \paren {\dfrac {\dfrac 1 2} {1!} } \paren {\dfrac 1 n + 1} + \sum_{k = 2}^\infty\frac{B_k}{k!}\paren{\paren{\frac 1 x }^\paren{k-1}(n) -(-1)^k\paren{\frac 1 x }^\paren{k-1}(1) }
      | c = Definition of natural logarithm, Natural Logarithm of 1 is 0, Definition:Bernoulli Numbers/Sequence
}}
{{eqn | r = \map \ln n + \paren {\dfrac 1 {2 n} + \dfrac 1 2} + \sum_{k = 2}^\infty\frac{B_k}{k!}\paren{\paren{\frac 1 x }^\paren{k-1}(n) -(-1)^k\paren{\frac 1 x }^\paren{k-1}(1) }
      | c = 
}}
{{eqn | r = \map \ln n + \paren {\dfrac 1 {2 n} + \dfrac 1 2} + \sum_{k = 1}^\infty\frac{B_{2k} }{\paren {2k}!}\paren{\paren{\frac 1 x }^\paren{2 k - 1}(n) - \paren{\frac 1 x }^\paren{2 k - 1}(1) }
      | c = Odd Bernoulli Numbers Vanish
}}
{{eqn | r = \map \ln n + \paren {\dfrac 1 {2 n} + \dfrac 1 2} + \paren {\dfrac {\dfrac 1 6} {2!} } \paren {-\dfrac 1 {n^2} + 1} + \sum_{k = 2}^\infty\frac{B_{2k} }{\paren {2k}!}\paren{\paren{\frac 1 x }^\paren{2 k - 1}(n) - \paren{\frac 1 x }^\paren{2 k - 1}(1) }
      | c = Power Rule for Derivatives
}}
{{eqn | r = \map \ln n + \paren {\dfrac 1 {2 n} - \dfrac 1 {12 n^2} } + \paren {\dfrac 1 2 + \dfrac 1 {12} } + \sum_{k = 2}^\infty\frac{B_{2k} }{\paren {2k}!}\paren{\paren{\frac 1 x }^\paren{2 k - 1}(n) - \paren{\frac 1 x }^\paren{2 k - 1}(1) }
      | c = reorganizing
}}
{{eqn | r = \map \ln n + \paren {\dfrac 1 {2 n} - \dfrac 1 {12 n^2} } + \paren {\dfrac 1 2 + \dfrac 1 {12} } + \paren {\dfrac {-\dfrac 1 {30} } {4!} } \paren {-\dfrac 6 {n^4} + 6} + \sum_{k = 3}^\infty\frac{B_{2k} }{\paren {2k}!}\paren{\paren{\frac 1 x }^\paren{2 k - 1}(n) - \paren{\frac 1 x }^\paren{2 k - 1}(1) }
      | c = {{Corollary|Power Rule for Derivatives}}
}}
{{eqn | r = \map \ln n + \paren {\dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} } + \paren {\dfrac 1 2 + \dfrac 1 {12} - \dfrac 1 {120}  } + \sum_{k = 3}^\infty\frac{B_{2k} }{\paren {2k}!}\paren{\paren{\frac 1 x }^\paren{2 k - 1}(n) - \paren{\frac 1 x }^\paren{2 k - 1}(1) }
      | c = reorganizing
}}
{{eqn | r = \map \ln n + \paren {\dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} } + \paren {\dfrac 1 2 + \dfrac 1 {12} - \dfrac 1 {120}  } + \paren {\dfrac {\dfrac 1 {42} } {6!} } \paren {-\dfrac {120} {n^6} + 120} + \sum_{k = 4}^\infty\frac{B_{2k} }{\paren {2k}!}\paren{\paren{\frac 1 x }^\paren{2 k - 1}(n) - \paren{\frac 1 x }^\paren{2 k - 1}(1) }
      | c = {{Corollary|Power Rule for Derivatives}}
}}
{{eqn | r = \map \ln n + \paren {\dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - \dfrac 1 {252 n^6} } + \paren {\dfrac 1 2 + \dfrac 1 {12} - \dfrac 1 {120} + \dfrac 1 {252} } + \sum_{k = 4}^\infty\frac{B_{2k} }{\paren {2k}!}\paren{\paren{\frac 1 x }^\paren{2 k - 1}(n) - \paren{\frac 1 x }^\paren{2 k - 1}(1) }
      | c = reorganizing
}}
{{end-eqn}}
From the {{Defof|Euler-Mascheroni Constant}}, we have
{{begin-eqn}}
{{eqn | l = \gamma
      | o = :=
      | r = \lim_{n \mathop \to +\infty} \paren {H_n - \ln n}
}}
{{end-eqn}}
where $\sequence {H_n}$ are the harmonic numbers and $\ln$ is the natural logarithm.
Taking the limit at infinity in our equation above, we observe:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to +\infty} \paren {\dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - \dfrac 1 {252 n^6} + \cdots }
      | r = 0
      | c = 
}}
{{end-eqn}}
Therefore, the sum of constants must be equal to $\gamma$
Therefore:
:$H_n \approx \ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - \epsilon$
:$0 < \epsilon < \dfrac 1 {252 n^6}$
{{qed}}