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

Question

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.

Answer 1

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 $\Delta x^{\underline m}=mx^{\underline{m-1}}$.