proof $ \binom{n+k+1}{k+1} \times \left( \frac{1}{\binom{n+k}{k}}- \frac{1}{\binom{n+k+1}{k}} \right)=\frac{k}{k+1} $ I would appreciate if somebody could help me with the following problem
Q: How to proof (by combinatorial proof)
$$ \binom{n+k+1}{k+1} \times \left( \frac{1}{\binom{n+k}{k}}
-  \frac{1}{\binom{n+k+1}{k}}  \right)=\frac{k}{k+1} $$
M1: I study (not combinatorial proof)
$$\binom{n+k+1}{k+1} \times \left( \frac{1}{\binom{n+k}{k}}
-  \frac{1}{\binom{n+k+1}{k}}  \right)=\frac{(n+k+1_)!}{(k+1)!n!}\times \left(\frac{k!n!}{(n+k)!}-\frac{k!(n+1)!}{(n+k+1)!} \right)
=\frac{n+k+1}{k+1}-\frac{n+1}{k+1}=\frac{k}{k+1}$$
I  wonder combinatorial proof!!!
 A: It’s a bit involved, but here is a combinatorial argument.
As usual, for $m\in\Bbb Z^+$ let $[m]=\{1,\ldots,m\}$. Let $\mathscr{A}$ be the family of $k$-element subsets of $[n+k+1]$, $\mathscr{B}$ the family of $(k+1)$-element subsets of $[n+k+1]$, $\mathscr{C}$ the family of $(k+1)$-element subsets of $[n+k+1]$ that contain the integer $n+k+1$, and $\mathscr{D}=\mathscr{B}\setminus\mathscr{C}$, the family of $(k+1)$-element subsets of $[n+k+1]$ that do not contain the integer $n+k+1$.
Each $A\in\mathscr{A}$ can be expanded to a member of $\mathscr{B}$ in $n+1$ different ways, but each member of $\mathscr{B}$ contains $\binom{k+1}k=k+1$ different members of $\mathscr{A}$, so
$$\binom{n+k+1}{k+1}=|\mathscr{B}|=\frac{n+1}{k+1}|\mathscr{A}|=\frac{n+1}{k+1}\binom{n+k+1}k\;,$$
and hence
$$\frac{\binom{n+k+1}{k+1}}{\binom{n+k+1}k}=\frac{n+1}{k+1}\;.$$
Now consider a set $C\in\mathscr{C}$. From $C$ we can form $n$ different members of $\mathscr{D}$ by replacing the integer $n+k+1$ with one of the $n$ elements of $[n+k]\setminus C$. However, each of the resulting members of $\mathscr{D}$ can be produced from any of $\binom{k+1}k=k+1$ members of $\mathscr{C}$, one for each of its $k$-element subsets, so
$$|\mathscr{D}|=\frac{n}{k+1}|\mathscr{C}|\;,$$
and
$$\binom{n+k+1}{k+1}=|\mathscr{B}|=|\mathscr{C}|+|\mathscr{D}|=|\mathscr{C}|\left(1+\frac{n}{k+1}\right)=\binom{n+k}k\left(1+\frac{n}{k+1}\right)\;,$$
and hence
$$\frac{\binom{n+k+1}{k+1}}{\binom{n+k}k}=1+\frac{n}{k+1}=\frac{n+k+1}{k+1}\;.$$
Thus,
$$\begin{align*}
\binom{n+k+1}{k+1}\left(\frac1{\binom{n+k}k}-\frac1{\binom{n+k+1}k}\right)&=\frac{\binom{n+k+1}{k+1}}{\binom{n+k}k}-\frac{\binom{n+k+1}{k+1}}{\binom{n+k+1}k}\\\\
&=\frac{n+k+1}{k+1}-\frac{n+1}{k+1}\\\\
&=\frac{k}{k+1}\;.
\end{align*}$$
