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

\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}
