Speaking in terms of Formal Series (as already noted in another answer) it is easy, starting from the known Power series $$ {{z^n } \over {\left( {1 - z} \right)^{n + 1} }} = \sum\limits_{0\, \le \,k} {\left( \matrix{ k \cr n \cr} \right)} \;z^k \quad {\rm integer n} \ge {\rm 0 } $$ to demonstrate that $$ \eqalign{ & \sum\limits_{0\, \le \,n} {a_{\,n} {{z^n } \over {\left( {1 - z} \right)^{n + 1} }}} = {1 \over {\left( {1 - z} \right)}}\sum\limits_{0\, \le \,n} {a_{\,n} \left( {{z \over {1 - z}}} \right)^n } = {1 \over {\left( {1 - z} \right)}}f(z) = \cr & = \sum\limits_{0\, \le \,n} {a_{\,n} \sum\limits_{0\, \le \,k} {\left( \matrix{ k \cr n \cr} \right)} \;z^k } = \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{0\, \le \,n} {\left( \matrix{ k \cr n \cr} \right)a_{\,n} } } \right)\;z^k } = \cr & = \sum\limits_{0\, \le \,k} {s_{\,k} \;z^k } = g(z) \cr} $$ and$$ \eqalign{ & \sum\limits_{0\, \le \,n} {a_{\,n} {{z^n } \over {\left( {1 - z} \right)^{n + 1} }}} = {1 \over {\left( {1 - z} \right)}}\sum\limits_{0\, \le \,n} {a_{\,n} \left( {{z \over {1 - z}}} \right)^n } = {1 \over {\left( {1 - z} \right)}}f\left( {{z \over {1 - z}}} \right) = \cr & = \sum\limits_{0\, \le \,n} {a_{\,n} \sum\limits_{0\, \le \,k} {\left( \matrix{ k \cr n \cr} \right)} \;z^k } = \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{0\, \le \,n} {\left( \matrix{ k \cr n \cr} \right)a_{\,n} } } \right)\;z^k } = \cr & = \sum\limits_{0\, \le \,k} {s_{\,k} \;z^k } = g(z) \cr} $$
and further variations by introducing $(-1)^n$ and $(-1)^{n-k}$