How to show $ \sum\limits_{n=0}^{a_1} \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2} $ identity? How do i show this identity:
$ \sum\limits_{n=0}^{a_1}  \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2}  = \sum\limits_{n=0}^{a_1}  \dfrac{a_1 \choose n}{a_1+a_2 \choose n} $
I don't think the partial sums  match  $  \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2}  ≠   \dfrac{a_1 \choose n}{a_1+a_2 \choose n}$
I was trying to understand why $\sum ^{a_{1}+a_{2}}_{n=1}\dfrac {\begin{pmatrix} a_{1} \\ n \end{pmatrix}}{\begin{pmatrix} a_{1}+a_{2} \\ n \end{pmatrix}}=\dfrac {a_{1}}{1+a_{2}}$, 
Somebody kindly outlined the  proof, using $\sum\limits_{m=0}^b  {c+m \choose c} ={b+c+1 \choose c+1} = \frac{b+c+1}{c+1} {b+c \choose c}$ as $\sum\limits_{n=1}^{a_1+a_2}  \dfrac{a_1 \choose n}{a_1+a_2 \choose n} = \sum\limits_{n=0}^{a_1}  \dfrac{a_1 \choose n}{a_1+a_2 \choose n} - 1$ = $= \sum\limits_{n=0}^{a_1}  \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2} - 1$ = $= \dfrac{a_2+a_1+1}{a_2+1}-1$
I understand apart from that identity.
 A: 
We start with the right-hand side and  obtain
  \begin{align*}
\color{blue}{\sum_{n=0}^{a_1}}&\color{blue}{\binom{a_1}{n}\binom{a_1+a_2}{n}^{-1}}\\
&=\sum_{n=0}^{a_1}\frac{a_1!}{n!(a_1-n)!}\cdot\frac{n!(a_1+a_2-n)!}{(a_1+a_2)!}\\
&=\frac{a_1!}{(a_1+a_2)!}\sum_{n=0}^{a_1}\frac{(a_1+a_2-n)!}{(a_1-n)!}\\
&=\frac{a_1!a_2!}{(a_1+a_2)!}\sum_{n=0}^{a_1}\frac{(a_2+n)!}{n!a_2!}\tag{1}\\
&\,\,\color{blue}{=\binom{a_1+a_2}{a_2}^{-1}\sum_{n=0}^{a_1}\binom{a_2+n}{a_2}}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we change the order of summation $n\to a_1-n$ and expand with $a_2!$.



The other identity
  \begin{align*}
\sum_{n=1}^{a_1}\binom{a_1}{n}\binom{a_1+a_2}{n}^{-1}=\frac{a_1}{1+a_2}
\end{align*}
  can be derived for instance using the Beta function with the identity
  \begin{align*}
\binom{p}{q}^{-1}=(p+1)\int_0^1x^q(1-x)^{p-q}\,dx
\end{align*}
We obtain
  \begin{align*}
\color{blue}{\sum_{n=1}^{a_1}}&\color{blue}{\binom{a_1}{n}\binom{a_1+a_2}{n}^{-1}}\\
&=(a_1+a_2+1)\int_0^1\sum_{n=1}^{a_1}\binom{a_1}{n}x^n(1-x)^{a_1+a_2-n}\,dx\\
&=(a_1+a_2+1)\int_0^1(1-x)^{a_1+a_2}\sum_{n=1}^{a_1}\binom{a_1}{n}\left(\frac{x}{1-x}\right)^n\,dx\\
&=(a_1+a_2+1)\int_0^1(1-x)^{a_1+a_2}\left[\left(1+\frac{x}{1-x}\right)^{a_1}-1\right]\,dx\\
&=(a_1+a_2+1)\int_0^1\left[(1-x)^{a_2}-(1-x)^{a_1+a_2}\right]\,dx\\
&=(a_1+a_2+1)\left[\frac{1}{a_2+1}-\frac{1}{a_1+a_2+1}\right]\\
&\,\,\color{blue}{=\frac{a_1}{a_2+1}}
\end{align*}
  and the claim follows.

