Using generating functions to solve binomial identities I would like to solve the two identities,
$$\sum_{j=0}^m \binom{n+j-1}{n-1} = \binom{n+m}{n}$$
$$\sum_{j=0}^m (-1)^{m-j} \binom{n+1}{m-j} \cdot \binom{n+j}{n} = 0$$
I was able to convince myself they are true using other methods, and now I am interested in a derivation using generating functions. How might I approach these problems?
 A: We can do these using coefficient extractors which uses generating
functions and sometimes, complex variables. For the first one we find
$$\sum_{j\ge 0} {n+j-1\choose n-1} [[j\le m]]
= \sum_{j\ge 0} {n+j-1\choose n-1} [z^m] \frac{z^j}{1-z}
\\ = [z^m] \frac{1}{1-z} \sum_{j\ge 0} {n+j-1\choose n-1} z^j
= [z^m] \frac{1}{1-z} \frac{1}{(1-z)^{n}}
\\ = [z^m] \frac{1}{(1-z)^{n+1}}
= {n+m\choose n}.$$
We get for the second one
$$[z^m] (1+z)^{n+1} \sum_{j=0}^m (-1)^{m-j} z^j {n+j\choose n}.$$
Here the coefficient extractor enforces the upper range of the sum and we have
$$[z^m] (1+z)^{n+1} \sum_{j\ge 0} (-1)^{m-j} z^j {n+j\choose n}
\\ = (-1)^m [z^m] (1+z)^{n+1} \frac{1}{(1+z)^{n+1}}
= (-1)^m [z^m] 1 = 0.$$
This is for $m\ge 1.$ We learn at this point that we needed neither                        residues nor complex variables.
A: For the first
$$
\eqalign{
  & \sum\limits_{0 \le m} {\sum\limits_{j = 0}^m {\left( \matrix{
  n + j - 1 \cr 
  n - 1 \cr}  \right)x^{\,j}\, y^{\,m} } }  =   \cr 
  & \sum\limits_{0 \le m} {\sum\limits_{j = 0}^m {\left( \matrix{
  n + j - 1 \cr 
  j \cr}  \right)x^{\,j} \, y^{\,m} \, } }  =   \cr 
  &  = \sum\limits_{0 \le j} {\sum\limits_{j \le m} {\left( \matrix{
   - n \cr 
  j \cr}  \right)\left( { - x} \right)^{\,j} \, y^{\,m} } }  =   \cr 
  &  = \sum\limits_{0 \le j} {\sum\limits_{0 \le m - j} {\left( \matrix{
   - n \cr 
  j \cr}  \right)\left( { - xy} \right)^{\,j} \, y^{\,m - j} } }  =   \cr 
  &  = {1 \over {\left( {1 - xy} \right)^{\,n} \left( {1 - y} \right)}}\;\;\buildrel {x = 1} \over
 \longrightarrow \;\;{1 \over {\left( {1 - y} \right)^{\,n + 1} }} =   \cr 
  &  = \sum\limits_{0 \le m} {\left( \matrix{
  n + m \cr 
  n \cr}  \right)y^{\,m} }  = \sum\limits_{0 \le m} {\left( \matrix{
  n + m \cr 
  m \cr}  \right)y^{\,m} }  = \sum\limits_{0 \le m} {\left( \matrix{
   - n - 1 \cr 
  m \cr}  \right)\left( { - y} \right)^{\,m} }   \cr 
  &  \cr} 
$$
where:

*

*

*

*symmetry ($0 \le n+j-1$);



*


*upper negation ($0 \le n+j-1$);



*


*change summation index ( $m \to m-j$)



*


*sums are disjoint, and put $x=1$



*


*sum over $y^m$ on the RHS.



Same track (more or less) for the second
$$
\eqalign{
  & \sum\limits_{0 \le m} {\sum\limits_{j = 0}^m {\left( { - 1} \right)^{\,m - j} \left( \matrix{
  n + 1 \cr 
  n - j \cr}  \right)\left( \matrix{
  n + j \cr 
  n \cr}  \right)x^{\,j} \, y^{\,m} \, } }  =   \cr 
  &  = \sum\limits_{0 \le m} {\sum\limits_{j = 0}^m {\left( { - 1} \right)^{\,j} \left( \matrix{
  n + 1 \cr 
  n - \left( {m - j} \right) \cr}  \right)\left( \matrix{
  n + m - j \cr 
  m - j \cr}  \right)x^{\,m - j} \, y^{\,m} \, } }  =   \cr 
  &  = \sum\limits_{0 \le m} {\sum\limits_{0 \le m - j} {\left( { - 1} \right)^{\,j} \left( \matrix{
  n + 1 \cr 
  n - \left( {m - j} \right) \cr}  \right)\left( \matrix{
  n + m - j \cr 
  m - j \cr}  \right)x^{\,m - j} \, y^{\,m} \, } }  =   \cr 
  &  = \sum\limits_{0 \le m} {\sum\limits_{0 \le k} {\left( { - 1} \right)^{\,m} \left( \matrix{
  n + 1 \cr 
  n - k \cr}  \right)\left( \matrix{
   - n - 1 \cr 
  k \cr}  \right)x^{\,k} \, y^{\,m} \, } } \;\buildrel {x = 1} \over
 \longrightarrow \;  \cr 
  &  \to \left( \matrix{
  0 \cr 
  n\, \cr}  \right){1 \over {\left( {1 + y} \right)}} = \delta _{n,0}  \cr} 
$$
