# 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.

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.

• @ Yes, step (1) was what I could not get, changing the order of summation back to front. Thank you very much, it seems a very long winded way to prove $\sum ^{a_{1}+a_{2}}_{n=1}\dfrac {\begin{pmatrix} a_{1} \\ n \end{pmatrix}}{\begin{pmatrix} a_{1}+a_{2} \\ n \end{pmatrix}}=\dfrac {a_{1}}{1+a_{2}}$ – Tinatim Jun 27 '18 at 13:50
• Of course, my apologies. Ok, what's the straight-forward approach then? – Tinatim Jun 27 '18 at 14:28
• @Tinatim: Everything's fine. Proof of the other identity added. Regards, – Markus Scheuer Jun 27 '18 at 14:51
• Nice, I assume this should read \begin{align*} \binom{p}{q}^{-1}=(p+1)\int_0^1x^q(1-x)^{p-q}\,dx \end{align*} I will look at the beta function. – Tinatim Jun 27 '18 at 16:56
• @Tinatim: This approach is often useful when working with reciprocals of binomial coefficients. Thanks for pointing at the mistake. Corrected. – Markus Scheuer Jun 27 '18 at 17:03