Alternating sum of binomial coefficients multiplied by index to an n-2 extent

Question

Doing research in probability modelling I obtained an identity, which is correct for $n > 2$.

$$\sum\limits^n_{i=1}(-1)^{n+i}{{n}\choose{i}}i^{n-2}=0$$

How can it be proven directly? In what literature can I find it?

Answer 1

Let write the required series as $$ S_n=\sum_{i=1}^{n} (-1)^i~ i^{n-2} ~ {n\choose i}\tag 1$$ Let us consider an interesting function $$f(x)=(1-e^{x})^n = 1~-~n e^{x}+\frac{n(n-1)}{2!} e^{2x}-\frac {n(n-1)(n-2)}{3!} e^{3x}+....(n+1)~ \mbox{terms}\tag 2$$ Next the coefficient of $x^{n-2}$ in the RHS of $(2)$ is $$-\frac{n}{1!} \frac{1^{n-2}}{(n-2)!}+\frac{n(n-1)}{2!} \frac{2^{n-2}}{(n-2)!}-\frac{n(n-1)(n-2)}{3!} \frac{3^{n-2}}{(n-2)!}+....=\frac{S_n}{(n-2)!} \tag 3$$ Also, the coefficient of $x^{n-2}$ in $f(x)$ can be obtained from $$f(x)=(-1)^n (e^x-1)^n = (-1)^n \left ( x^n+\frac{n}{2}x^{n+1}+...\right),$$ where there are terms having higher powers of $x$, namely $x^n$ on wards. It therefore follows that $S_n=0.$

Answer 2

This is a variation using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. Recalling $e^z=\sum_{j=0}^\infty \frac{z^j}{j!}$ we can write for instance \begin{align*} n![z^n]e^{kz}=k^n\tag{1} \end{align*}

We obtain for $n>2$ \begin{align*} \color{blue}{\sum_{k=1}^n}&\color{blue}{(-1)^kk^{n-2}\binom{n}{k}}\\ &=\sum_{k=1}^n(-1)^k\binom{n}{k}(n-2)![z^{n-2}]e^{kz}\tag{2}\\ &=(n-2)![z^{n-2}]\sum_{k=1}^n\binom{n}{k}\left(-e^{z}\right)^k\\ &=(n-2)![z^{n-2}]\left(\left(1-e^z\right)^n-1\right)\tag{3}\\ &=(n-2)![z^{n-2}]\left((-1)^n\left(z+\frac{z^2}{2}+\cdots\right)^n-1\right)\tag{4}\\ &\,\,\color{blue}{=0} \end{align*}

Comment:

• In (2) we apply the coefficient of operator according to (1)

• In (3) we apply the binomial theorem.

• In (4) we observe the coefficient of $z^{n-2}$ is zero since the left term starts with powers in $z$ greater or equal to $n$ and the constant $-1$ does not contribute since $n>2$.

Hint: Instructive examples using this and related techniques can be found in H.S. Wilf's book generatingfunctionology. See the examples following (1.2.6).

Answer 3

My favorite proof of the probably most general form of the identity starts with the claim: $$\frac{d^m}{dx^m}(e^x-1)^n=\sum_{k=0}^m {m \brace k}\frac{n!}{(n-k)!} (e^x-1)^{n-k}e^{kx},\tag1 $$ which can be easily proved by induction over $m$ using the well-known summation property of the Stirling numbers of the second kind: $$ {m \brace k-1}+k{m \brace k}={m+1 \brace k}. $$

Now consider the obvious equality: $$ (e^x-1)^n=\sum_{k=0}^n (-1)^{n-k}\binom nk e^{kx}.\tag2 $$ Differentiating both sides $m$ times with respect to $x$ and substituting $x=0$ in the resulting expression: $$\sum_{k=0}^m {m \brace k}\frac{n!}{(n-k)!} (e^x-1)^{n-k}e^{kx}=\sum_{k=0}^n (-1)^{n-k}\binom nk k^me^{kx},\tag3 $$ one obtains: $$ \boxed{{m \brace n}n!=\sum_{k=0}^n (-1)^{n-k}\binom nk k^m}\tag4 $$ since only $k=n$ can provide a non-zero term in the left hand side sum upon the substitution. Particularly for $m<n$ the sum is $0$ and your claim follows.

Answer 4

Just for the sake of somebody giving an answer not using the exponential function (for a question that does not involve it), let me take a shot at this one. I'll voluntarily also refrain from invoking finite differences (as this was already mentioned in a comment to the question), although they obviously inspired some of the arguments I will give.

A key observation is the obvious fact that the sequence $(i^{n-2})_{i\in\Bbb N}$ lies in the space of sequences of polynomial growth of degree less than$~n$ (i.e., of sequences $(P[i])_{i\in\Bbb N}$ for such a polynomial). A slightly modified version of the statement will remain true if we replace the factor $i^{n-2}$ by the corresponding term of any such sequence. The modification consists of extending the summation by starting at $i=0$ rather than $i=1$, which makes no difference for the given expression, but does make a difference after mentioned replacement.

It is well known (and quite obvious) that the "columns" of Pascal's triangle (Pascal's own term was "rangs perpendiculaires") are sequences of polynomial growth of degree equal to the index of column. Concretely, I'll take as column $k$ the sequence $(\binom{k+i}k)_{i\in\Bbb N}$ (so I don't include any terms $0$ from before it meets Pascal's triangle proper), and writing $\binom{k+i}k=\frac{(i+1)(i+2)\ldots(i+k)}{k!}$ makes clear this is a sequence of polynomial growth of degree$~k$. It follows easily that these columns for $k=0,1,\ldots,n-1$ form a basis of the space of sequences of polynomial growth of degree less than$~n$. Therefore I can prove my claim by showing $$ \sum_{i=0}^n(-1)^{n-i}\binom ni\binom{k+i}k=0 \qquad\text{for $k=0,1,\ldots,n-1$}\tag{1} $$ (I've used $(-1)^{n+i}=(-1)^{n-i}$). I will actually show, with the convention $\binom nm=0$ whenever $m<0$, that $$ \sum_{i=0}^n(-1)^{n-i}\binom n{n-i}\binom{k+i}k=\binom k{k-n} \qquad\text{for all $k,n\in\Bbb N$}\tag{2} $$ which clearly gives $(1)$ as a special case (now I've rewritten $\binom ni=\binom n{n-i}$ for future convenience).

One proof of $(2)$ uses a few algebraic manipulations and binomial coefficients with negative upper index. The RHS can be written as $\binom k{k-n}=\binom kn=(-1)^n\binom{n-k-1}n$ where the final equality is the "negation of the upper index" operation. By the same token the factor $\binom{k+i}k$ in the LHS equals $(-1)^i\binom{-k-1}i$, and $(2)$ simplifies to $$ \sum_{i=0}^n\binom n{n-i}\binom{-k-1}i=\binom{n-k-1}n \qquad\text{for all $k,n\in\Bbb N$,}\tag{3} $$ which is just an instance of the Vandermonde identity (which remains valid for binomial coefficients with negative upper indices).

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Another way to prove $(2)$ is to use induction on $n$, but this requires generalising the statement first (so-called induction loading), or at least this greatly simplifies the task. The generalisation consists in replacing two instances of $k$ by an independent new parameter $d$ (which could be taken to live in $\Bbb Z$ or even be an indeterminate, but I'll stick to plain binomial coefficients with indices in $\Bbb N$ here): $$ \sum_{i=0}^n(-1)^{n-i}\binom n{n-i}\binom{d+i}k=\binom d{k-n} \qquad\text{for all $k,n,d\in\Bbb N$}\tag{4} $$ The proof is now straightforward, so I'll just sketch it (doing this for yourself is more instructive than seeing it done anyway). The base case $n=0$ being easy, assume $n>0$ and rewrite $\binom n{n-i} = \binom{n-1}{n-i} + \binom{n-1}{n-i-1}$ (Pascal's recurrence), and split the summation accordingly into two summations; in the (second) summation with $\binom{n-1}{n-i-1}$, the term $i=n$ can be dropped, and what remains gives $-\binom{d}{k-(n-1)}$ by the induction hypothesis; in the (first) summation with $\binom{n-1}{n-i}$, the term $i=0$ can be dropped, and after then shifting down the summation index by$~1$, the induction hypothesis again applies and makes the sum $\binom{d+1}{k-(n-1)}$; so one gets $\binom{d+1}{k-(n-1)}-\binom{d}{k-(n-1)} = \binom d{k-n}$, again Pascal's recurrence.

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Finally let me attempt a bijective proof of $(4)$, just for fun, knowing that what I will propose is not really natural enough to afford an "explanation" any better then the given proofs. For $m\in\Bbb N$, let $[m]$ be the set of the first $m$ natural numbers. The collection I will consider is that of pairs $(S,w)$ where $S\subseteq[n]$ and $w$ is a string of $d+i$ bits in $\{0,1\}$ with sum $k$, where $i$ is the number of elements in the complement of$~S$. I want to match up most such $(S,w)$ two by two, in such a way that the parity of$~i$ is different between each matching pair, and such that the remaining pairs have $S=\emptyset$ (so $i=n$, which ensures a positive sign) and gives rise bijectively to a $k-n$ element subset of $[d]$. Starting both in $[n]$ and $w$ at the left, repeat the following. If for the next element $j$ of $[n]$ one has $j\in S$, a matching $(S',w')$ is produced by setting $S'=S\setminus\{j\}$, while $w'$ is obtained from$~w$ by inserting a bit $0$ at the current position to form $w'$, and stop; otherwise inspect the next bit of $w$, and if it is $0$, a matching $(S',w')$ is produced by removing this bit $0$ to form $w'$ while setting $S'=S\cup\{j\}$, and stop. In the remaining case (both $j\notin S$ and $w$ had a bit $1$), advance $j$ if possible and move across the considered bit $1$ of $w$, and repeat. If this terminates without producing a match, one had $S=\emptyset$ and $w$ had length $d+n$ and starts with $n$ bits $1$; the remaining string of $w$ encodes a $k-n$ element subset of $[d]$ as desired. (In the case $n>k$ that was our original interest, this final case can never be reached as $w$ has only $k$ bits $1$ to begin with.)