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

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## Contents

## 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:

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

$\blacksquare$

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $39$