# 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:

$\displaystyle \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$

where:

$\displaystyle {n \brack k}$ 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 $\map P n$ be the proposition:

$\displaystyle \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$

$\map P 0$ is the case:

 $\displaystyle \sum_k \paren {-1}^k {0 \brack k}$ $=$ $\displaystyle \sum_k \delta_{0 k}$ 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 $\map P 0$ is seen to hold.

$\map P 1$ is the case:

 $\displaystyle \sum_k \paren {-1}^k {1 \brack k}$ $=$ $\displaystyle \sum_k \paren {-1}^k \delta_{1 k}$ 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 $\map P 1$ is seen to hold.

### Basis for the Induction

$\map P 2$ is the case:

 $\displaystyle \sum_k \paren {-1}^k {2 \brack k}$ $=$ $\displaystyle - {2 \brack 1} + {2 \brack 2}$ Definition of Summation $\displaystyle$ $=$ $\displaystyle - {2 \brack 1} + 1$ 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 $\map P 2$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

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

So this is the induction hypothesis:

$\displaystyle \sum_k \paren {-1}^k {m \brack k} = \delta_{m 0} - \delta_{m 1} = 0$

from which it is to be shown that:

$\displaystyle \sum_k \paren {-1}^k {m + 1 \brack k} = \delta_{\paren {m + 1} 0} - \delta_{\paren {m + 1} 1} = 0$

### Induction Step

This is the induction step:

 $\displaystyle \sum_k \paren {-1}^k {m + 1 \brack k}$ $=$ $\displaystyle \sum_k \paren {-1}^k \paren {m {m \brack k} + {m \brack k - 1} }$ Definition of Unsigned Stirling Numbers of the First Kind $\displaystyle$ $=$ $\displaystyle m \sum_k \paren {-1}^k {m \brack k} + \sum_k \paren {-1}^k {m \brack k - 1}$ $\displaystyle$ $=$ $\displaystyle m \sum_k \paren {-1}^k {m \brack k} - \sum_k \paren {-1}^k {m \brack k}$ Translation of Index Variable of Summation $\displaystyle$ $=$ $\displaystyle \paren {m - 1} \sum_k {m \brack k}$ $\displaystyle$ $=$ $\displaystyle \paren {m + 1} \times 0$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle 0$

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

Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$

$\blacksquare$