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I tried evaluating the integral $\int\limits_{[0,1]^n} \min(x^1,x^2,\ldots,x^n) \lvert d^nx\rvert$.

In my first attempt, I used a recursive approach and managed to defined the integral as being $I_n^k = \int\limits_{[0,1]^n} \min(x^1,x^2,\ldots,x^n)^k \lvert d^nx\rvert$, where I could evaluate

$I_n^k = I_{n-1}^ k - \frac{k}{k+1} \cdot I_{n-1}^{k+1}$, and

$I_1^k = \frac{1}{k+1}$.

Using this recursive approach, I managed to extract the

$I_n^1 = \sum\limits_{k=0}^{n-1} {n-1 \choose k}\cdot \frac{(-1)^k}{(k+1)(k+2)}$

I plugged numbers into this sum for multiple values of $n$, and saw that I constantly get $\frac{1}{n+1}$.

During my second attempt, I managed to eventually find the integral by splitting the area into $n!$ areas in which there exists some order for each element $x_0$ such that $x_0^{i_1} \leq x_0^{i_2} \leq \ldots \leq x_0^{i_n}$, and proved that the integral over each of these area is $\frac{1}{(n+1)!}$ which gave me showed me more definitively that the integral is equal to $\frac{1}{n+1}$.

That said, after trying for a while, I couldn't come up with any combinatorial/algebraic proof that the sum I found is indeed $\frac{1}{n+1}$. I tried evaluating it as a telescoping sum, giving me the expression $\sum_{k=0}^{\frac{n}{2}}{n \choose 2k}\cdot\frac{\left(4k+3-n\right)}{\left(2k+3\right)\left(2k+2\right)\left(2k+1\right)}$, but couldn't expand this sum to anything useful either. I haven't worked much with sums of this form and was wondering whether I'm missing something that can help me show this without the integral.

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  • $\begingroup$ You can show that $\frac{(1+x)^{n+2}-1-(n+2)x}{(n+1)(n+2)}=\sum_{k=0}^n {n\choose k}\frac{ x^{k+2}}{(k+1)(k+2)}$ by using the rules of differentiation of polynomials and the binomial theorem (ie. no analysis). $\endgroup$
    – reuns
    Jan 19, 2022 at 9:10

7 Answers 7

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Since I'm currently studying Discrete Mathematics, I present to you a proof using tools from this domain. More precisely, we could make use of the following result:

Difference formula: If $f(x)$ is defined for all integers $x$, then $$\Delta^nf(x)=\sum_{k=0}^n\binom nk (-1)^{n-k}f(x+k).$$

Here, $\Delta f(x)=f(x+1)-f(x)$, and $\Delta^n$ is the application of the $\Delta$-operator $n$ times.

Choosing $f(x)=\frac 1{(x+1)(x+2)}=x^{\underline {-2}}$ ($x$ to the $-2$ falling) and $x=0$, the RHS equals $$(-1)^n\sum_{k=0}^n\binom nk (-1)^k\frac{1}{(k+1)(k+2)}.$$

Thus, your sum equals the LHS multiplied by $(-1)^n$, that is $$(-1)^n\Delta^nf(0)=(-1)^n\frac{(-1)^n(n+1)!}{(n+2)!}=\frac{1}{n+2}.$$

I made use of the fact that $\Delta x^{\underline m}=mx^{\underline{m-1}}$.

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\begin{aligned}\sum_{k=0}^{n}{\binom{n}{k}\frac{\left(-1\right)^{k}}{\left(k+1\right)\left(k+2\right)}}&=\sum_{k=0}^{n}{\binom{n}{k}\int_{-1}^{0}{x^{k}\left(1+x\right)\mathrm{d}x}}\\ &=\int_{-1}^{0}{\left(1+x\right)\sum_{k=0}^{n}{\binom{n}{k}x^{k}}\,\mathrm{d}x}\\ &=\int_{-1}^{0}{\left(1+x\right)\left(1+x\right)^{n}\,\mathrm{d}x}\\ &=\left[\frac{\left(1+x\right)^{n+2}}{n+2}\right]_{-1}^{0}\\ &=\frac{1}{n+2}\end{aligned}

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Proof that $$\sum\limits_{k=0}^{n} {n \choose k}\cdot \frac{(-1)^k}{(k+1)(k+2)} = \frac{1}{n+2} .$$

Begin with the binomial theorem $$ \sum_{k=0}^n {n \choose k} x^k = (1+x)^n . $$ Integrate $\int_0^y$: $$ \sum_{k=0}^n {n \choose k} \frac{y^{k+1}}{k+1} = \frac{(1+y)^{n+1}-1}{n+1} . $$ Integrate $\int_0^z$: $$ \sum_{k=0}^n {n \choose k} \frac{z^{k+2}}{(k+1)(k+2)} = \frac{(1+z)^{n+2}-1-(n+2)z}{(n+1)(n+2)} . $$ Substitute $z=-1$: $$ \sum_{k=0}^n {n \choose k} \frac{(-1)^{k}}{(k+1)(k+2)} = \frac{n+1}{(n+1)(n+2)} = \frac{1}{n+2} . $$


Perhaps the same proof shows $$ \sum\limits_{k=0}^{\infty} {\alpha \choose k}\cdot \frac{(-1)^k}{(k+1)(k+2)} = \frac{1}{\alpha+2} $$ for real (or complex) $\alpha \ne -2$.

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    $\begingroup$ Your last sentence inspired me. It works like a charm ! Cheers & thanks :-) $\endgroup$ Jan 19, 2022 at 10:01
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After @GEdgar's answer, inspired by the approach, a generalization is $$\sum\limits_{k=0}^{\infty} {a \choose k}\, \frac{(-1)^k}{(k+b)(k+c)} =\frac{\Gamma (a+1)}{c-b} \left(\frac{\Gamma (b)}{\Gamma (a+b+1)}-\frac{\Gamma (c)}{\Gamma (a+c+1)}\right)$$

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Multiplying both sides by $(n+1)(n+2)$ yields the equivalent identity $$\sum_{k=0}^n (-1)^k \binom{n+2}{k+2} = n+1,$$ which follows directly from the well-known identity $$\sum_{j=0}^m (-1)^j \binom{m}{j} = 0$$ by taking $m=n+2$ and stripping off the first two terms.

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This answer comes a bit late but I thought it is worth mentioning it because it shows a simple algebraic proof of the formula in question.

First note the following two facts:

  • (1): $\frac 1{(k+1)(k+2)}\binom nk = \frac 1{(n+1)(n+2)}\binom{n+2}{k+2}$, Indeed \begin{eqnarray*} \frac 1{(k+1)(k+2)}\binom nk & = & \frac{n!}{k!(k+1)(k+2)(n-k)!} \\ & = & \frac 1{(n+1)(n+2)}\frac{(n+2)!}{(k+2)!((n+2)-(k+2))!} \\ & = & \frac 1{(n+1)(n+2)}\binom{n+2}{k+2} \end{eqnarray*}
  • (2): According to the binomial formula, we have $$0 = (1-1)^{n+2} = \sum_{k=0}^{n+2}(-1)^k\binom{n+2}{k}$$

Now, we have \begin{eqnarray*} \sum\limits_{k=0}^{n} {n \choose k}\cdot \frac{(-1)^k}{(k+1)(k+2)} & \stackrel{(1)}{=} & \frac 1{(n+1)(n+2)} \sum\limits_{\color{blue}{k=0}}^{n} (-1)^k \cdot \binom{n+2}{k+2}\\ & \stackrel{(-1)^{k+2}=(-1)^k}{=} & \frac 1{(n+1)(n+2)} \sum\limits_{\color{blue}{k=2}}^{n+2} (-1)^k \cdot \binom{n+2}{k} \\ & \stackrel{(2)}{=} & \frac 1{(n+1)(n+2)}\left(0 - \binom{n+2}{0} + \binom{n+2}{1}\right) \\ & = & \frac{n+1}{(n+1)(n+2)} = \frac 1{n+2} \end{eqnarray*}

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A variation based upon telescoping.

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{n}{k}\frac{(-1)^k}{(k+1)(k+2)}}\\ &=\sum_{k=0}^n\binom{n}{k}(-1)^k\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\\ &=\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{1}{k+1}-\sum_{k=1}^{n+1}\binom{n}{k-1}(-1)^{k-1}\frac{1}{k+1}\tag{1}\\ &=\sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^k\frac{1}{k+1}\tag{2}\\ &=\frac{1}{n+2}\sum_{k=0}^{n+1}\binom{n+2}{k+1}(-1)^k\tag{3}\\ &=\frac{-1}{n+2}\sum_{k=1}^{n+2}\binom{n+2}{k}(-1)^{k}\tag{4}\\ &=\frac{-1}{n+2}(1-1)^{n+2}+\frac{1}{n+2}\tag{5}\\ &\,\,\color{blue}{=\frac{1}{n+2}} \end{align*} and the claim follows.

Comment:

  • In (1) we shift the index of the right-hand sum to start with $k=1$.

  • In (2) we use $\binom{p}{q}+\binom{p}{q-1}=\binom{p+1}{q}$.

  • In (3) we use $\frac{p+1}{q+1}\binom{p}{q}=\binom{p+1}{q+1}$.

  • In (4) we shift the index to start with $k=1$.

  • In (5) we apply the binomial theorem.

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