Summation Formula for Reciprocal of Binomial Coefficient/Proof 2

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Theorem

\(\ds \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x}\) \(=\) \(\ds \dfrac 1 {x \binom {n + x} n}\)
\(\ds \) \(=\) \(\ds \dfrac {n!} {x \paren {x + 1} \cdots \paren {x + n} }\)

as long as the denominators are not zero.


Proof

Consider the value of $\map \Beta {x, n + 1}$, where $\Beta$ is the beta function.

We have:

\(\ds \map \Beta {x, n + 1}\) \(=\) \(\ds \int_0^1 t^{x - 1} \paren {1 - t}^n \rd t\) Definition 1 of Beta Function
\(\ds \) \(=\) \(\ds \int_0^1 t^{x - 1} \sum_{k \mathop \ge 0} \binom n k \paren {-t}^k 1^{n - k} \rd t\) Binomial Theorem
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \binom n k \paren {-1}^k \int_0^1 t^{x - 1} t^k \rd t\) Linear Combination of Definite Integrals
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \binom n k \paren {-1}^k \int_0^1 t^{x + k - 1} \rd t\) Index Laws/Sum of Indices
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \binom n k \paren {-1}^k \frac 1 {x + k}\)

which equates to the left hand side of the identity.

We also have:

\(\ds \map \Beta {x, n + 1}\) \(=\) \(\ds \frac {\map \Gamma x \map \Gamma {n + 1} } {\map \Gamma {x + n + 1} }\) Definition 3 of Beta Function
\(\ds \) \(=\) \(\ds \frac {n! \map \Gamma x } {\map \Gamma {x + n + 1} }\) Gamma Function Extends Factorial
\(\ds \) \(=\) \(\ds \frac {n! \map \Gamma x } {\paren {x + n} \map \Gamma {x + n} }\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds \frac {n! \map \Gamma x } {\paren {x + n - 1} \paren {x + n} \map \Gamma {x + n - 1} }\) Gamma Difference Equation
\(\ds \) \(:\) \(\ds \)
\(\ds \) \(=\) \(\ds \frac {n! \map \Gamma x } {x \paren {x + 1} \cdots \paren {x + n} \map \Gamma x}\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds \frac {n!} {x \paren {x + 1} \cdots \paren {x + n} }\)

which equates to the right hand side of the identity.

Hence the result.

$\blacksquare$


Sources