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

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Theorem

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

Then:

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

where:

$\ds {n \brack k}$ 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:

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


$\map P 0$ is the case:

\(\ds \sum_k \paren {-1}^k {0 \brack k}\) \(=\) \(\ds \sum_k \delta_{0 k}\) Unsigned Stirling Number of the First Kind of 0
\(\ds \) \(=\) \(\ds 1\) all terms vanish but for $k = 0$
\(\ds \) \(=\) \(\ds \delta_{0 0} - \delta_{0 1}\) Definition of Kronecker Delta

Thus $\map P 0$ is seen to hold.


$\map P 1$ is the case:

\(\ds \sum_k \paren {-1}^k {1 \brack k}\) \(=\) \(\ds \sum_k \paren {-1}^k \delta_{1 k}\) Unsigned Stirling Number of the First Kind of 1
\(\ds \) \(=\) \(\ds -1\) all terms vanish but for $k = 1$
\(\ds \) \(=\) \(\ds \delta_{1 0} - \delta_{1 1}\) Definition of Kronecker Delta

Thus $\map P 1$ is seen to hold.


Basis for the Induction

$\map P 2$ is the case:

\(\ds \sum_k \paren {-1}^k {2 \brack k}\) \(=\) \(\ds - {2 \brack 1} + {2 \brack 2}\) Definition of Summation
\(\ds \) \(=\) \(\ds - {2 \brack 1} + 1\) Unsigned Stirling Number of the First Kind of Number with Self
\(\ds \) \(=\) \(\ds -\binom 2 2 + 1\) Unsigned Stirling Number of the First Kind of n with n-1
\(\ds \) \(=\) \(\ds -1 + 1\) Binomial Coefficient with Self
\(\ds \) \(=\) \(\ds 0\) Binomial Coefficient with Self
\(\ds \) \(=\) \(\ds \delta_{2 0} - \delta_{2 1}\) 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:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

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


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


Therefore:

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

$\blacksquare$


Sources

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