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I want to show that \begin{align} &\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{k+n+1}=\frac{n!\,m!}{(n+m+1)!}\\ \end{align}

I came across this when trying to prove
\begin{align} \int_0^1 (1-x)^m x^n dx =\frac{n!\,m!}{(n+m+1)!}\\ \end{align} My teacher proved it by using the substitution $x=\sin^2 (t)$ and then using the Wallis formula but I tried to prove it by using Binomial Theorem and integrating each term and I got stuck at this step.

Jean-Claude Arbaut
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Anshuman Agrawal
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    It's not difficult to expand the integrand using the binomial theorem, and then integrate, to get the sum in your first line. However, you still need to integrate $(1-x)^mx^n$ directly. Try repeated integration by parts to reduce one of the exponents until it vanishes: it's another way to prove your second line. Note that this integral is also called the Beta function. – Jean-Claude Arbaut Jan 13 '22 at 07:46
  • You can also try to write the integral in terms of the Beta and Gamma functions. – F_M_ Jan 13 '22 at 07:48
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    @Jean-Claude Arbaut Thanks for the help! I managed to get the result by repeated integration by parts and in fact see that the method is quite similar to the way I had proved the Wallis formula in class from the formula of the integral sin^n(x) from 0 to 1, I also had a rough idea that this way could be used to prove this result – Anshuman Agrawal Jan 13 '22 at 08:05
  • I'd still like to see a direct way to prove this result though – Anshuman Agrawal Jan 13 '22 at 08:11
  • Please do not use displays in titles. – Pedro Jan 13 '22 at 08:38
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    See here, here, here, here, here, here. Links found through Approach Zero. – Fabius Wiesner Jan 13 '22 at 14:56
  • @Pedro Tamaroff I'm terribly sorry for that, I don't how to write in code and I put together this answer by copying the code from a few answers – Anshuman Agrawal Jan 13 '22 at 15:05

3 Answers3

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Let $k\in\mathbb{N} $, we have the following : \begin{aligned}I_{n+k,m-k}=\int_{0}^{1}{x^{n+k}\left(1-x\right)^{m-k}\,\mathrm{d}x}&=\left[\frac{x^{n+k+1}\left(1-x\right)^{m-k}}{n+k+1}\right]_{0}^{1}+\frac{m-k}{n+k+1}\int_{0}^{1}{x^{n+k+1}\left(1-x\right)^{m-k-1}\,\mathrm{d}x}\\ I_{n+k,m-k}&=\frac{m-k}{n+k+1}I_{n+k+1,m-k-1}\\ \Longrightarrow\prod_{k=0}^{m-1}{\frac{I_{n+k,m-k}}{I_{n+k+1,m-k-1}}}&=\prod_{k=0}^{m-1}{\frac{m-k}{n+k+1}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{I_{n,m}}{I_{n+m,0}}&=\prod_{k=0}^{m-1}{\frac{m-k}{n+k+1}}\\ I_{n,m}&=\frac{n!m!}{\left(n+m+1\right)!}\end{aligned}

CHAMSI
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There is the “Linear algebra” way. Let $E$ be the space of functions from $(0,\infty)$ to $\Bbb R$ and define the operator $S\colon f\mapsto Sf$ on $E$ by $$Sf(x):=f(x+1),\quad f\in E,\ x>0.$$ Let the difference operator $\Delta:=\mathrm{Id}_E-S$. Since $\mathrm{Id}_E$ and $-S$ commute, the Binomial theorem gives you $\Delta^m$ for any integer $m\ge0$: $$\Delta^m=\sum_{k=0}^m\binom mk(\mathrm{Id}_E)^{m-k}(-S)^k,$$ that is $$\Delta^mf(x)=\sum_{k=0}^m\binom mk(-1)^kf(x+k).\tag{$\star$}$$ Now consider $f\in E$ given by $f(x):=\frac1x$. We can see that \begin{align*} \Delta f(x)&=\frac1x-\frac1{x+1}=\frac1{x(x+1)},\\[.4em] \Delta^2f(x):=\Delta(\Delta f)(x)&=\frac1{x(x+1)}-\frac1{(x+1)(x+2)}=\frac2{x(x+1)(x+2)}, \end{align*} and an immediate induction shows that $$\Delta^m f(x)=\frac{m!}{x(x+1)\cdots(x+m)}.$$ Thus $(\star)$ reads $$\frac{m!}{x(x+1)\cdots(x+m)}=\sum_{k=0}^m\binom mk(-1)^k\frac1{x+k}.$$ This is an identity between functions of the variable $x>0$ and it suffices to take $x:=n+1$ to establish your claim.

nejimban
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In trying to prove

$$S_m = \sum_{k=0}^m \frac{(-1)^k}{k+n+1} {m\choose k} = \frac{n! m!}{(n+m+1)!}$$

we introduce

$$f(z) = m! (-1)^m \frac{1}{z+n+1} \prod_{q=0}^m \frac{1}{z-q}.$$

This has the property that for $0\le k\le m$

$$\mathrm{Res}_{z=k} f(z) = m! (-1)^m \frac{1}{k+n+1} \prod_{q=0}^{k-1} \frac{1}{k-q} \prod_{q=k+1}^m \frac{1}{k-q} \\ = m! (-1)^m \frac{1}{k+n+1} \frac{1}{k!} \frac{(-1)^{m-k}}{(m-k)!} = \frac{(-1)^k}{k+n+1} {m\choose k}.$$

Note that the residue of $f(z)$ at infinity is zero by inspection and residues sum to zero so that we have

$$S_m = \sum_{k=0}^m \mathrm{Res}_{z=k} f(z) = - \mathrm{Res}_{z=-n-1} f(z) \\ = - \mathrm{Res}_{z=-n-1} \; m! (-1)^m \frac{1}{z+n+1} \prod_{q=0}^m \frac{1}{z-q} = - m! (-1)^m \prod_{q=0}^m \frac{1}{-n-1-q} \\ = m! \prod_{q=0}^m \frac{1}{n+1+q} = \frac{m! n!}{(n+m+1)!}.$$

This is the claim.

Marko Riedel
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