Stirling Numbers Proof

Question

Let $n > 1$ be an integer. Prove the following: $$\sum\limits_{k=1}^{\infty} (-1)^k (k - 1)! S(n,k) = 0$$ where $S(n,k)$ is a Stirling number of the second kind.

(Hint: Recurrence Relation)

Workings:

The recurrence relation of Stirling numbers of the second kind I believe is:

$S(n+1,k) = k S(n,k) + S(n,k-1)$

Though I do not see how this will potentially help out.

Any help will be appreciated.

Answer 1

If you use the hint and let

$$ b_k = k! \,S(n-1,k)+(k-1)!S(n-1,k-1) = a_k+a_{k-1} $$

then the series becomes

$$ \sum\limits_{k=1}^{∞} (−1)^k b_k = \sum\limits_{k=1}^{∞} (−1)^k (a_k+a_{k-1}) $$

and you will notice that when you expand the series terms will cancel each other.

Answer 2

Since the recursion gives: $$ x^n = \sum_{k=1}^{n} S(n,k)(x)_k = \sum_{k=1}^{n} S(n,k)x(x-1)\cdot\ldots\cdot(x-(k-1)) $$ (see also line $(11)$ here) we have that: $$x^{n-1}= \sum_{k=1}^{n} S(n,k)(x-1)\cdot\ldots\cdot(x-(k-1))\tag{1}$$ and the claim simply follows from evaluating the previous identity in $x=0$.

Answer 3

Seeing that the OGF of the Stirling numbers of the second kind has already been used I would like to point out that this can be done using the EGF as well. Recall the species of set partitions which is $$\mathfrak{P}(\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ which gives the bivariate generating function $$G(z,u) = \exp(u(\exp(z)-1)).$$ It follows immediately that $${n\brace k} = n![z^n] \frac{(\exp(z)-1)^k}{k!}.$$

Suppose we seek to evaluate $$\sum_{k=1}^n (-1)^k\times (k-1)! \times {n\brace k}.$$

Substitute the EGF value into the sum to get $$n! [z^n] \sum_{k=1}^n (-1)^k \times (k-1)! \times \frac{(\exp(z)-1)^k}{k!}$$ which is $$n! [z^n] \sum_{k=1}^n (-1)^k \times \frac{(\exp(z)-1)^k}{k}.$$ Now observe that $(\exp(z)-1)^k$ starts at the power $z^k$ so we may extend the sum to infinity without affecting the count, getting $$n! [z^n] \sum_{k\ge 1} (-1)^k \times \frac{(\exp(z)-1)^k}{k} = n! [z^n] \log\frac{1}{1+(\exp(z)-1)} \\ = n! [z^n] \log \exp(-z) = n! [z^n] (-z).$$ This is $-1$ when $n=1$ and zero otherwise.