Showing $\sum\limits_{j=0}^M \frac{M \choose j}{N+M \choose j} = \frac{N+M+1}{N+1}$ 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?
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\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}
