Summation over Lower Index of Unsigned Stirling Numbers of the First Kind with Alternating Signs

Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:

$\displaystyle \sum_k \left({-1}\right)^k \left[{n \atop k}\right] = \delta_{n 0} - \delta_{n 1}$

where:

$\displaystyle \left[{n \atop k}\right]$ denotes an unsigned Stirling number of the first kind
$\delta_{n 0}$ denotes the Kronecker delta.

Proof

The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \sum_k \left({-1}\right)^k \left[{n \atop k}\right] = \delta_{n 0} - \delta_{n 1}$

$P \left({0}\right)$ is the case:

 $\displaystyle \sum_k \left({-1}\right)^k \left[{0 \atop k}\right]$ $=$ $\displaystyle \sum_k \delta_{0 k}$ Unsigned Stirling Number of the First Kind of 0 $\displaystyle$ $=$ $\displaystyle 1$ all terms vanish but for $k = 0$ $\displaystyle$ $=$ $\displaystyle \delta_{0 0} - \delta_{0 1}$ Definition of Kronecker Delta

Thus $P \left({0}\right)$ is seen to hold.

$P \left({1}\right)$ is the case:

 $\displaystyle \sum_k \left({-1}\right)^k \left[{1 \atop k}\right]$ $=$ $\displaystyle \sum_k \left({-1}\right)^k \delta_{1 k}$ Unsigned Stirling Number of the First Kind of 1 $\displaystyle$ $=$ $\displaystyle -1$ all terms vanish but for $k = 1$ $\displaystyle$ $=$ $\displaystyle \delta_{1 0} - \delta_{1 1}$ Definition of Kronecker Delta

Thus $P \left({1}\right)$ is seen to hold.

Basis for the Induction

$P \left({2}\right)$ is the case:

 $\displaystyle \sum_k \left({-1}\right)^k \left[{2 \atop k}\right]$ $=$ $\displaystyle -\left[{2 \atop 1}\right] + \left[{2 \atop 2}\right]$ Definition of Summation $\displaystyle$ $=$ $\displaystyle -\left[{2 \atop 1}\right] + 1$ Unsigned Stirling Number of the First Kind of Number with Self $\displaystyle$ $=$ $\displaystyle -\binom 2 2 + 1$ Unsigned Stirling Number of the First Kind of n with n-1 $\displaystyle$ $=$ $\displaystyle -1 + 1$ Binomial Coefficient with Self $\displaystyle$ $=$ $\displaystyle 0$ Binomial Coefficient with Self $\displaystyle$ $=$ $\displaystyle \delta_{2 0} - \delta_{2 1}$ Definition of Kronecker Delta

Thus $P \left({2}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis

Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 2$, then it logically follows that $P \left({m + 1}\right)$ is true.

So this is the induction hypothesis:

$\displaystyle \sum_k \left({-1}\right)^k \left[{m \atop k}\right] = \delta_{m 0} - \delta_{m 1} = 0$

from which it is to be shown that:

$\displaystyle \sum_k \left({-1}\right)^k \left[{m + 1 \atop k}\right] = \delta_{\left({m + 1}\right) 0} - \delta_{\left({m + 1}\right) 1} = 0$

Induction Step

This is the induction step:

 $\displaystyle \sum_k \left({-1}\right)^k \left[{m + 1 \atop k}\right]$ $=$ $\displaystyle \sum_k \left({-1}\right)^k \left({m \left[{m \atop k}\right] + \left[{m \atop k - 1}\right]}\right)$ Definition of Unsigned Stirling Numbers of the First Kind $\displaystyle$ $=$ $\displaystyle m \sum_k \left({-1}\right)^k \left[{m \atop k}\right] + \sum_k \left({-1}\right)^k \left[{m \atop k - 1}\right]$ $\displaystyle$ $=$ $\displaystyle m \sum_k \left({-1}\right)^k \left[{m \atop k}\right] - \sum_k \left({-1}\right)^k \left[{m \atop k}\right]$ Translation of Index Variable of Summation $\displaystyle$ $=$ $\displaystyle \left({m - 1}\right) \sum_k \left[{m \atop k}\right]$ $\displaystyle$ $=$ $\displaystyle \left({m + 1}\right) \times 0$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle 0$

So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \left({-1}\right)^k \left[{n \atop k}\right] = \delta_{n 0} - \delta_{n 1}$

$\blacksquare$