Showing $\sum\limits_{j=0}^M \frac{M \choose j}{N+M \choose j} = \frac{N+M+1}{N+1}$

Question

In an answer to another question, I stated $$\sum\limits_{j=0}^M \frac{M \choose j}{N+M \choose j} = \frac{N+M+1}{N+1}.$$

It is clearly true when $N=0$ since you add up $M+1$ copies of $1$, and when $M=0$ since you add up one copy of $1$. And, for example, with $M=4$ and $N=9$ you get $\frac{1}{1}+\frac{4}{13}+\frac{6}{78}+\frac{4}{286}+\frac{1}{715} = \frac{14}{10}$ as expected.

But how might you approach a general proof?

Answer 1

Hint. Note that $$\frac{M \choose j}{N+M \choose j}=\frac{\binom{N+M-j}{N}}{\binom{N+M}{N}}.$$ Hence, we can rewrite the sum as $$\sum_{j=0}^M \frac{M \choose j}{N+M \choose j}=\frac{1}{\binom{N+M}{N}}\sum_{j=0}^M \binom{N+M-j}{N}=\frac{1}{\binom{N+M}{N}}\sum_{i=N}^{N+M} \binom{i}{N}.$$ Finally use the Hockey-stick identity.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{j = 0}^{M}{{M \choose j} \over {N + M \choose j}} = {N + M + 1 \over N + 1}:\ {\LARGE ?}}$.

\begin{align} \sum_{j = 0}^{M}{{M \choose j} \over {N + M \choose j}} & = \sum_{j = 0}^{M}{M!/\bracks{j!\pars{M - j}!} \over \pars{N + M}!/\bracks{j!\pars{N + M - j}!}} \\[5mm] & = {M!\, N! \over \pars{N + M}!}\sum_{j = 0}^{M} {N + M - j \choose M - j} \\[5mm] & = {M!\, N! \over \pars{N + M}!}\pars{-1}^{M} \sum_{j = 0}^{M}\pars{-1}^{j} \bracks{z^{M - j}}\pars{1 + z}^{-N - 1} \\[5mm] & = {M!\, N! \over \pars{N + M}!}\pars{-1}^{M} \bracks{z^{M}}\pars{1 + z}^{-N - 1}\, {\pars{-z}^{M + 1} - 1 \over \pars{-z} - 1} \\[5mm] & = {M!\, N! \over \pars{N + M}!}\pars{-1}^{M} \bracks{z^{M}}\pars{1 + z}^{-N - 2} \\[5mm] & = {M!\, N! \over \pars{N + M}!}\pars{-1}^{M} \braces{{-\bracks{-N - 2} + M - 1 \choose M}\pars{-1}^{M}} \\[5mm] & = \bbx{N + M + 1 \over N + 1} \end{align}