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

Theorem
Let $x \in \R_{> 0}$ be a real number.

Then:


 * $\displaystyle \sum_{k \mathop \in \Z} \binom n k x^k H_k = \left({x + 1}\right)^n \left({H_n - \ln \left({1 + \frac 1 x}\right)}\right) + \epsilon$

where:
 * $\dbinom n k$ denotes a binomial coefficient
 * $H_k$ denotes the $k$th harmonic number
 * $0 < \epsilon < \dfrac 1 {x \left({n + 1}\right)}$

Proof
Let $S_n := \displaystyle \sum_{k \mathop \in \Z} \binom n k x^k H_k$.

Then:

As $S_1 = x$, we have that:


 * $\dfrac {S_n} {\left({x + 1}\right)^n} = H_n - \displaystyle \sum_{k \mathop = 1}^n \frac 1 {k \left({x + 1}\right)^k}$

From Power Series Expansion for Logarithm of 1 over 1-z we have:
 * $\ln \dfrac 1 {1 - z} = \displaystyle \sum_{k \mathop \ge 1} \frac {z^k} k$

Thus:

So as $n \to \infty$, $\displaystyle \sum_{k \mathop \ge 1} \frac 1 {k \left({x + 1}\right)^k}$ converges, and so:

The result follows.