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

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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

$\blacksquare$


Sources