Theorem: Sum over k of n Choose k by x to the k by kth Harmonic Number

Let $x \in \R_{> 0}$ be a real number.
Then:
:$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \paren {x + 1}^n \paren {H_n - \map \ln {1 + \frac 1 x} } + \epsilon$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$H_k$ denotes the $k$th harmonic number
:$0 < \epsilon < \dfrac 1 {x \paren {n + 1} }$

Proof:
Let $S_n := \ds \sum_{k \mathop \in \Z} \binom n k x^k H_k$.
Then:
{{begin-eqn}}
{{eqn | l = S_{n + 1}
      | r = \sum_{k \mathop \in \Z} \binom {n + 1} k x^k H_k
      | c = 
}}
{{eqn | r = \sum_{k \mathop \in \Z} \paren {\binom n k + \binom n {k - 1} } x^k H_k
      | c = Pascal's Rule
}}
{{eqn | r = \sum_{k \mathop \in \Z} \binom n k x^k H_k + \sum_{k \mathop \in \Z} \binom n {k - 1} x^k H_k
      | c = 
}}
{{eqn | r = S_n + x \sum_{k \mathop \ge 1} \binom n {k - 1} x^{k - 1} \paren {H_{k - 1} + \frac 1 k}
      | c = Definition of $S_n$
}}
{{eqn | r = S_n + x \sum_{k \mathop \ge 1} \binom n {k - 1} x^{k - 1} H_{k - 1} + x \sum_{k \mathop \ge 1} \binom n {k - 1} x^{k - 1} \frac 1 k
      | c = 
}}
{{eqn | r = S_n + x \sum_{k \mathop \ge 0} \binom n k x^k H_k + \sum_{k \mathop \ge 1} \binom n {k - 1} x^k \frac 1 k
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = S_n + x S_n + \sum_{k \mathop \ge 1} \frac k {n + 1} \binom {n + 1} k x^k \frac 1 k
      | c = Definition of $S_n$, Factors of Binomial Coefficient
}}
{{eqn | r = \paren {x + 1} S_n + \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k x^k
      | c = simplifying
}}
{{eqn | r = \paren {x + 1} S_n + \frac 1 {n + 1} \paren {\sum_{k \mathop \ge 0} \binom {n + 1} k x^k - \binom {n + 1} 0 x^0}
      | c = forcing the summation to include the $k = 0$ term
}}
{{eqn | r = \paren {x + 1} S_n + \frac {\paren {x + 1}^{n + 1} - 1} {n + 1}
      | c = Binomial Theorem and Binomial Coefficient with $0$
}}
{{eqn | ll= \leadsto
      | l = \frac {S_{n + 1} } {\paren {x + 1}^{n + 1} }
      | r = \frac {S_n} {\paren {x + 1}^n} + \frac 1 {n + 1} - \frac 1 {\paren {n + 1} \paren {x + 1}^{n + 1} }
      | c = dividing through by $\paren {x + 1}^{n + 1}$
}}
{{end-eqn}}
As $S_1 = x$, we have that:
{{begin-eqn}}
{{eqn | l = \dfrac {S_n} {\paren {x + 1}^n}
      | r = \frac {S_{n - 1} } {\paren {x + 1}^{n - 1} } + \frac 1 n - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = \frac {S_{n - 2} } {\paren {x + 1}^{n - 2} } + \frac 1 {n - 1} - \frac 1 {\paren {n - 1} \paren {x + 1}^{n - 1} } + \frac 1 n - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = \dots}}
{{eqn | r = \frac {S_1} {\paren {x + 1}^1} + \frac 1 2 + \frac 1 3 + \dots + \frac 1 n - \frac 1 {2 \paren {x + 1}^2} - \frac 1 {3 \paren {x + 1}^3} - \dots - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = \frac x {x + 1} - 1 + \frac 1 {x + 1} + 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 n - \frac 1 {x + 1} - \frac 1 {2 \paren {x + 1}^2} - \frac 1 {3 \paren {x + 1}^3} - \dots - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 n - \frac 1 {x + 1} - \frac 1 {2 \paren {x + 1}^2} - \frac 1 {3 \paren {x + 1}^3} - \dots - \frac 1 {n \paren {x + 1}^n}
}}
{{eqn | r = H_n - \ds \sum_{k \mathop = 1}^n \frac 1 {k \paren {x + 1}^k}
      | c = {{Defof|Harmonic Number}}
}}
{{end-eqn}}
From Power Series Expansion for $\map \ln {1 - x}$ we have:
:$\ds \map \ln {1 - z} = -\sum_{k \mathop \ge 1} \frac {z^k} k$
Hence by Logarithm of Reciprocal we have:
:$\ds \ln \frac 1 {1 - z} = -\map \ln {1 - z} = \sum_{k \mathop \ge 1} \frac {z^k} k$
Thus:
{{begin-eqn}}
{{eqn | l = \map \ln {1 + \frac 1 x}
      | r = \ln \frac 1 {1 - \frac 1 {x + 1} }
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 1} \frac 1 {k \paren {x + 1}^k}
      | c = 
}}
{{end-eqn}}
So as $n \to \infty$, $\ds \sum_{k \mathop \ge 1} \frac 1 {k \paren {x + 1}^k}$ converges, and so:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \frac 1 {k \paren {x + 1}^k}
      | o = <
      | r = \frac 1 {\paren {n + 1} \paren {x + 1}^{n + 1} } \sum_{k \mathop \ge 0} \frac 1 {\paren {x + 1}^k}
      | c = 
}}
{{eqn | r = \frac 1 {\paren {n + 1} \paren {x + 1}^n x}
      | c = 
}}
{{end-eqn}}
The result follows.
{{qed}}