First note that: {{begin-eqn}} {{eqn | o =        | r = \sum_{k \mathop \in \Z} \binom n k \dfrac {\paren {-1}^k k} {k + x} + \sum_{k \mathop \in \Z} \binom n k \dfrac {\paren {-1}^k x} {k + x}       | c =  }} {{eqn | r = \sum_{k \mathop \in \Z} \binom n k \paren {-1}^k \dfrac {k + x} {k + x}       | c =  }} {{eqn | r = \sum_{k \mathop \in \Z} \paren {-1}^k \binom n k       | c =  }} {{eqn | r = 0       | c = Alternating Sum and Difference of Binomial Coefficients for Given n }} {{end-eqn}} The proof continues by induction on $n$. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$ === Basis for the Induction === $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \binom 0 k       | r = \binom 0 0 \dfrac {\paren {-1}^0} x       | c = Zero Choose n: $\dbinom 0 k = 0$ for $k > 0$ }} {{eqn | r = \frac 1 x       | c =  }} {{eqn | r = \frac 1 {x \binom {0 + x} 0}       | c = Binomial Coefficient with Zero: $\dbinom {0 + x} 0 = 1$ }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P m$ is true, for all $m \ge 0$, then it logically follows that $\map P {m + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{k \mathop \ge 0} \binom m k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {m + x} m}$ from which it is to be shown that: :$\displaystyle \sum_{k \mathop \ge 0} \binom {m + 1} k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {m + 1 + x} {m + 1} }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \binom {m + 1} k \frac {\paren {-1}^k} {k + x}       | r = \sum_{k \mathop \ge 0} \binom m k \frac {\paren {-1}^k} {k + x} + \sum_{k \mathop \ge 0} \binom m {k - 1} \frac {\paren {-1}^k} {k + x}       | c = Pascal's Rule }} {{eqn | r = \sum_{k \mathop \ge 0} \binom m k \frac {\paren {-1}^k} {k + x} - \sum_{k \mathop \ge 0} \binom m {k - 1} \frac {\paren {-1}^{k - 1} } {k + x}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \binom m k \frac {\paren {-1}^k} {k + x} - \sum_{k \mathop \ge -1} \binom m k \frac {\paren {-1}^k } {k + 1 + x}       | c =  }} {{eqn | r = \frac 1 {x \binom {m + x} m} - \frac 1 {\paren {x + 1} \binom {m + x + 1} m} - \binom m {-1} \frac {\paren {-1}^{-1} } x       | c = Induction Hypothesis }} {{eqn | r = m! \paren {\frac 1 {x \paren {x + 1} \cdots \paren {x + m} } - \frac 1 {\paren {x + 1} \paren {x + 2} \cdots \paren {x + m + 1} } }       | c =  }} {{eqn | r = \frac {m!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + m} \paren {x + m + 1} } \paren {x + m + 1 - x}       | c =  }} {{eqn | r = \frac {\paren {m + 1}!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + m} \paren {x + m + 1} }       | c =  }} {{eqn | r = \frac 1 {x \binom {m + 1 + x} {m + 1} }       | c =  }} {{end-eqn}} So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \displaystyle \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$ {{qed}}