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I ran into an alternating sum in my research and would like to know if the following identity is true: $$ \sum_{i = 0}^{\left\lfloor \left(n + 1\right)/2\right\rfloor} \frac{\left(n + 1 - 2i\right)^{n + 1}}{2^{n}\left(n + 1\right)!}\binom{n + 1}{i}\left(-1\right)^{i} = 1\quad \forall\ \mbox{positive integers}\ n\geq 3 $$ Any help would be appreciated!.

Edit.
We might try to use an Iverson bracket $[[2q\le n]]$ in attempting to evaluate

$$S_n = \sum_{q=0}^{\lfloor n/2\rfloor} (n-2q)^n {n\choose q} (-1)^q.$$

We obtain

$$[v^n] \frac{1}{1-v} \sum_{q\ge 0} v^{2q} (n-2q)^n {n\choose q} (-1)^q$$

Using a coefficient extractor,

$$n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\; \frac{1}{v^{n+1}} \frac{1}{1-v} \sum_{q\ge 0} v^{2q} \exp(-2qz) {n\choose q} (-1)^q \\ = n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\; \frac{1}{v^{n+1}} \frac{1}{1-v} (1-v^2\exp(-2z))^n.$$

Now residues sum to zero and the residue at one yields

$$- n! [z^n] \exp(nz) (1-\exp(-2z))^n.$$

We have that since $(1-\exp(-2z))^n = (2z-2z^2\pm\cdots)^n = 2^n z^n +\cdots$ this evaluates to $-2^n n!.$ We find for the residue at infinity

$$- n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\; \frac{1}{v^2} v^{n+1} \frac{1}{1-1/v} (1-\exp(-2z)/v^2)^n \\ = n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\; \frac{1}{v^n} \frac{1}{1-v} (v^2-\exp(-2z))^n \\ = n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\; \frac{1}{v^n} \frac{1}{1-v} \sum_{q=0}^n {n\choose q} (-1)^{n-q} \exp(-2(n-q)z) v^{2q} \\ = \;\underset{v}{\mathrm{res}}\; \frac{1}{v^n} \frac{1}{1-v} \sum_{q=0}^n {n\choose q} (-1)^{n-q} (2q-n)^n v^{2q} \\ = \sum_{q=0}^n {n\choose q} (-1)^q (n-2q)^n [[2q\le n-1]].$$

Now when $n$ is odd this gives the upper limit $\lfloor n/2\rfloor$ and when $n$ is even $\lfloor n/2\rfloor -1$ however in the latter case we may raise to $\lfloor n/2\rfloor$ because the added term is zero in the sum per $(n-2q)^n = 0$. We have obtained

$$\sum_{q=0}^{\lfloor n/2\rfloor} {n\choose q} (-1)^q (n-2q)^n = S_n.$$

Collecting everything we have shown that $S_n - 2^n n! + S_n = 0$ or $S_n = 2^{n-1} n!.$ The question now becomes, is there a simpler proof?

RobPratt
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Wonmat
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    In which mathematical problem/theory such an ugly summation occured? – Bob Dobbs Sep 27 '22 at 14:13
  • I can confirm, it is true for $n=3$. So you can start with an induction. – Dietrich Burde Sep 27 '22 at 14:38
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    You could also show that $$ \sum_{i=0}^{n+1} \frac{(n+1-2i)^{n+1}}{2^n(n+1)!}\binom{n+1}{i}(-1)^i=2$$ or $$ \sum_{i=0}^{u} \frac{(u-2i)^{u}}{u!}\binom{u}{i}(-1)^i=2^u$$ – garondal Sep 27 '22 at 15:49
  • @DietrichBurde I thought about induction, but it seems that the -2i part is a bit annoying. – Wonmat Sep 27 '22 at 16:21
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    @BobDobbs It's a bit hard to include all the background here. I was doing some calculation with toric varieties, and this arises from a volume calculation, where I was doing the naive inclusion-exclusion principle. – Wonmat Sep 27 '22 at 16:25
  • Maybe this is helpful: https://math.stackexchange.com/questions/591350/expressing-a-factorial-as-difference-of-powers-sum-r-0n-binomnr-1 – garondal Sep 27 '22 at 16:46
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    "the $-2i$ part is a bit annoying" $j:=(n+1)/2-i$ is the way around that, at least for odd $n$. There's already a great answer, but it's often useful to reverse the order of terms with a new choice of dummy index. – J.G. Sep 27 '22 at 21:20
  • Creation/proof/disproof of. Result is Always interning thing irrespective of the fact wether it's has dome application. – Z Ahmed Oct 01 '22 at 03:16
  • Does this answer your question? proof of formula and calculation sum – RobPratt Oct 01 '22 at 04:32

1 Answers1

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We seek to show that

$$\sum_{q=0}^{\lfloor n/2\rfloor} (n-2q)^n {n\choose q} (-1)^q = 2^{n-1} n!.$$

Observe that

$$(n-2q)^n {n\choose q} (-1)^q = \frac{1}{2} \left[ (n-2q)^n {n\choose q} (-1)^q + (2q-n)^n {n\choose n-q} (-1)^{n-q} \right].$$

Hence we get for our sum

$$\frac{1}{2} \sum_{q=0}^{\lfloor n/2\rfloor} (n-2q)^n {n\choose q} (-1)^q + \frac{1}{2} \sum_{q=0}^{\lfloor n/2\rfloor} (2q-n)^n {n\choose n-q} (-1)^{n-q} \\ = \frac{1}{2} \sum_{q=0}^{\lfloor n/2\rfloor} (n-2q)^n {n\choose q} (-1)^q + \frac{1}{2} \sum_{q=n-\lfloor n/2\rfloor}^n (n-2q)^n {n\choose q} (-1)^q \\ = \frac{1}{2} \sum_{q=0}^n (n-2q)^n {n\choose q} (-1)^q.$$

Introducing a coefficient extractor,

$$\frac{1}{2} \sum_{q=0}^n {n\choose q} (-1)^q n! [z^n] \exp((n-2q) z) \\ = \frac{1}{2} n! [z^n] \exp(nz) \sum_{q=0}^n {n\choose q} (-1)^q \exp((-2q) z) \\ = \frac{1}{2} n! [z^n] \exp(nz) (1-\exp(-2z))^n.$$

Note however that $(1-\exp(-2z))^n = (2z-2z^2\pm\cdots)^n$ so the only contribution to the coefficient extractor $[z^n]$ is from the first term of the series so that $[z^n] \exp(nz) (2z-2z^2\pm\cdots)^n = 2^n$ and we finally have

$$2^{n-1} n!$$

as claimed.

Marko Riedel
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