Theorem: Summation Formula for Reciprocal of Binomial Coefficient

{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x}
      | r = \dfrac 1 {x \binom {n + x} n}
      | c = 
}}
{{eqn | r = \dfrac {n!} {x \paren {x + 1} \cdots \paren {x + n} }
      | c = 
}}
{{end-eqn}}
as long as the denominators are not zero.

Proof:
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:
:$\ds \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom 0 k \dfrac {\paren {-1}^k} {x + k}
      | r = \binom 0 0 \dfrac {\paren {-1}^0} {x + 0}
      | 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:
:$\ds \sum_{k \mathop \ge 0} \binom m k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {m + x} m}$
from which it is to be shown that:
:$\ds \sum_{k \mathop \ge 0} \binom {m + 1} k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {m + 1 + x} {m + 1} }$
=== 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:
:$\forall n \in \Z_{\ge 0}: \ds \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$
{{qed}}