# Sum over k of n Choose k by x to the k by kth Harmonic Number/x = -1

## Theorem

While for $x \in \R_{>0}$ be a real number:

$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \paren {x + 1}^n \paren {H_n - \map \ln {1 + \frac 1 x} } + \epsilon$

when $x = -1$ we have:

$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \dfrac {-1} n$

where:

$\dbinom n k$ denotes a binomial coefficient
$H_k$ denotes the $k$th harmonic number.

## Proof

When $x = -1$ we have that $1 + \dfrac 1 x = 0$, so $\map \ln {1 + \dfrac 1 x}$ is undefined.

Let $S_n = \ds \sum_{k \mathop \in \Z} \binom n k x^k H_k$

Then:

 $\ds S_{n + 1}$ $=$ $\ds \sum_{k \mathop \in \Z} \binom {n + 1} k x^k H_k$ $\ds$ $=$ $\ds \sum_{k \mathop \in \Z} \paren {\binom n k + \binom n {k - 1} } x^k H_k$ Pascal's Rule $\ds$ $=$ $\ds S_n + x \sum_{k \mathop \ge 1} \paren {\binom n {k - 1} } x^{k - 1} \paren {H_{k - 1} + \frac 1 k}$ $\ds$ $=$ $\ds S_n + x S_n + \sum_{k \mathop \ge 1} \binom n {k - 1} x^k \frac 1 k$ $\ds$ $=$ $\ds \paren {1 + x} S_n + \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k x^k$ (look up result for this)

Setting $x = -1$:

 $\ds S_{n + 1}$ $=$ $\ds \paren {1 + -1} S_n + \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k \paren {-1}^k$ $\ds$ $=$ $\ds \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k \paren {-1}^k$ $\ds \leadsto \ \$ $\ds S_n$ $=$ $\ds \frac 1 n \sum_{k \mathop \ge 1} \binom n k \paren {-1}^k$ $\ds$ $=$ $\ds \frac 1 n \paren {\sum_{k \mathop \ge 0} \binom n k \paren {-1}^k - \binom n 0}$ $\ds$ $=$ $\ds \frac 1 n \paren {0 - \binom n 0}$ Alternating Sum and Difference of Binomial Coefficients for Given n $\ds$ $=$ $\ds \frac {-1} n$ Binomial Coefficient with Zero

$\blacksquare$