Strange binomial sum

Question

Let $S_r(n) = \displaystyle\sum_{m=0}^n (-1)^m m^r \binom{n}{m}$, where $r$ is a non-negative integer. Show that $S_r(n) = 0$ for $r<n$.

My argument was along the lines of: $(1-z)^n = \displaystyle\sum_{m=0}^n \binom{n}{m}(-1)^m z^m$, so we differentiate this, multiply by $z$, repeat this $r$ times and then let $z=1$. As $r<n$ all terms on LHS have a factor of $(1-z)$ so all vanish and we're left with $0$.

I wasn't very satisfied with this argument and was wondering if there is a more algebraic approach which also doesn't let it become a horrible, unclear mess?

Answer 1

We have that

$$S_r(n) = r! [z^r] \sum_{m=0}^n (-1)^m \exp(mz) {n\choose m} \\ = r! [z^r] (1-\exp(z))^n.$$

Now $1-\exp(z) = - z - z^2/2 - z^3/6 -\cdots$ so $(1-\exp(z))^n = (-1)^n z^n + \cdots$. Hence we get zero from the coefficient extractor in $z$ when $r\lt n.$ This question has appeared here with some regularity.

Answer 2

Exponential Generating Function

Here is a different presentation along the lines of Marko Riedel's answer. $$ \begin{align} \sum_{r=0}^\infty\frac{x^r}{r!}\color{#C00}{S_r(n)} &=\sum_{r=0}^\infty\frac{x^r}{r!}\color{#C00}{\sum_{m=0}^n(-1)^mm^r\binom{n}{m}}\tag{1a}\\ &=\sum_{m=0}^n(-1)^me^{mx}\binom{n}{m}\tag{1b}\\[3pt] &=\left(1-e^x\right)^n\tag{1c}\\[6pt] &=x^n\left(\frac{1-e^x}x\right)^n\tag{1d} \end{align} $$ Comparing coefficients of $x^r$ between the left hand side of $\text{(1a)}$ and $\text{(1d)}$ shows that $S_r(n)=0$ for $r\lt n$.

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Induction

This can also be proven by induction.

For $r=0$ and $n\gt r$, the Binomial Theorem gives $$ \begin{align} S_0(n) &=\sum_{m=0}^n(-1)^m\binom{n}{m}\tag{2a}\\ &=(1-1)^n\tag{2b}\\[9pt] &=0\tag{2c} \end{align} $$ Assume that $S_k(n-1)=0$ for all $k\lt n-1$. Then, for $1\le r\lt n$, $$ \begin{align} S_r(n) &=\sum_{m=0}^n(-1)^mm^r\binom{n}{m}\tag{3a}\\ &=\sum_{m=1}^n(-1)^mm^r\frac nm\binom{n-1}{m-1}\tag{3b}\\ &=n\sum_{m=0}^{n-1}(-1)^{m+1}\color{#C00}{(m+1)^{r-1}}\binom{n-1}{m}\tag{3c}\\ &=-n\sum_{m=0}^{n-1}(-1)^m\color{#C00}{\sum_{k=0}^{r-1}m^k\binom{r-1}{k}}\binom{n-1}{m}\tag{3d}\\ &=-n\sum_{k=0}^{r-1}\binom{r-1}{k}S_k(n-1)\tag{3e}\\[9pt] &=0\tag{3f} \end{align} $$ Explanation:
$\text{(3a):}$ definition of $S$
$\text{(3b):}$ $r\ge1$ so we can leave out the $m=0$ term
$\phantom{\text{(3b):}}$ $\binom{n}{m}=\frac nm\binom{n-1}{m-1}$
$\text{(3c):}$ substitute $m\mapsto m+1$
$\text{(3d):}$ Binomial Theorem
$\text{(3e):}$ definition of $S$
$\text{(3f):}$ Since $k\le r-1\lt n-1$, we have $S_k(n-1)=0$ by assumption

Thus, by induction, $S_r(n)=0$ for all $r\lt n$.

Answer 3

For a combinatorial proof, interpret the sum as an inclusion-exclusion formula for a certain type of function, and observe that such functions do not exist when $r<n$.