Summation Formula for Reciprocal of Binomial Coefficient

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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 1

First note that:

\(\ds \) \(\) \(\ds \sum_{k \mathop \in \Z} \binom n k \dfrac {\paren {-1}^k k} {k + x} + \sum_{k \mathop \in \Z} \binom n k \dfrac {\paren {-1}^k x} {k + x}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in \Z} \binom n k \paren {-1}^k \dfrac {k + x} {k + x}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in \Z} \paren {-1}^k \binom n k\)
\(\ds \) \(=\) \(\ds 0\) Alternating Sum and Difference of Binomial Coefficients for Given n


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:

\(\ds \sum_{k \mathop \ge 0} \binom 0 k \dfrac {\paren {-1}^k} {x + k}\) \(=\) \(\ds \binom 0 0 \dfrac {\paren {-1}^0} {x + 0}\) Zero Choose n: $\dbinom 0 k = 0$ for $k > 0$
\(\ds \) \(=\) \(\ds \frac 1 x\)
\(\ds \) \(=\) \(\ds \frac 1 {x \binom {0 + x} 0}\) Binomial Coefficient with Zero: $\dbinom {0 + x} 0 = 1$


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:


\(\ds \sum_{k \mathop \ge 0} \binom {m + 1} k \frac {\paren {-1}^k} {k + x}\) \(=\) \(\ds \sum_{k \mathop \ge 0} \binom m k \frac {\paren {-1}^k} {k + x} + \sum_{k \mathop \ge 0} \binom m {k - 1} \frac {\paren {-1}^k} {k + x}\) Pascal's Rule
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \binom m k \frac {\paren {-1}^k} {k + x} - \sum_{k \mathop \ge 0} \binom m {k - 1} \frac {\paren {-1}^{k - 1} } {k + x}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \binom m k \frac {\paren {-1}^k} {k + x} - \sum_{k \mathop \ge -1} \binom m k \frac {\paren {-1}^k } {k + 1 + x}\)
\(\ds \) \(=\) \(\ds \frac 1 {x \binom {m + x} m} - \frac 1 {\paren {x + 1} \binom {m + x + 1} m} - \binom m {-1} \frac {\paren {-1}^{-1} } x\) Induction Hypothesis
\(\ds \) \(=\) \(\ds m! \paren {\frac 1 {x \paren {x + 1} \cdots \paren {x + m} } - \frac 1 {\paren {x + 1} \paren {x + 2} \cdots \paren {x + m + 1} } }\)
\(\ds \) \(=\) \(\ds \frac {m!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + m} \paren {x + m + 1} } \paren {x + m + 1 - x}\)
\(\ds \) \(=\) \(\ds \frac {\paren {m + 1}!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + m} \paren {x + m + 1} }\)
\(\ds \) \(=\) \(\ds \frac 1 {x \binom {m + 1 + x} {m + 1} }\)

So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 0}: \ds \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$

$\blacksquare$


Proof 2

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