0
$\begingroup$

I'm looking for a citable reference for the following identity involving the Stirling numbers of the second kind $S(n, k)$ stated in Equation (27): For $n \geq 2$, $$ \sum_{m=1}^n S(n, m) (-1)^m (m-1)! = 0. $$ Thank you.

$\endgroup$
3

1 Answer 1

5
$\begingroup$

The first formula in Section 24.1.4.I.B in Handbook of mathematical functions with formulas, graphs, and mathematical tables, Dover edition, 1965 (Library of Congress Catalog Card Number: 65.12253) by Abramowitz and Stegun is $$ x^n=\sum_{m=0}^n S(n,m) x(x-1)\cdots(x-m+1). \tag{1}\label{1} $$ Assuming here $n>1$ (so that $S(n,0)=0$), dividing both sides by $x$, and then letting $x=0$, we get the desired identity $$\sum_{m=1}^n S(n,m)(-1)^m(m-1)!=0.$$


One may note that Riordan (An Introduction to Combinatorial Analysis, Wiley, 1958, formula (35) on p. 33) uses identity \eqref{1} to define the Stirling numbers $S(n,m)$, then deriving their combinatorial definition (as the number of ways of partitioning a set of $n$ elements into $m$ nonempty sets) -- cf. formula (38) on p. 33 and the third display from the bottom on p. 91 of the book.

$\endgroup$
3
  • $\begingroup$ You can also substitute $x=-1$ $\endgroup$ Jan 19, 2023 at 18:49
  • 1
    $\begingroup$ @მამუკაჯიბლაძე : Thank you for your comment. Then you will probably need some other identity as well, because on the right-hand side you will have $\sum_{m=1}^n S(n,m)(-1)^m m!$ rather than $\sum_{m=1}^n S(n,m)(-1)^m(m-1)!$ (also $(-1)^n$ instead of $0$ on the left). $\endgroup$ Jan 19, 2023 at 18:59
  • $\begingroup$ Sorry for confusion, indeed this is a different identity! $\endgroup$ Jan 19, 2023 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.