Summation Formula for Reciprocal of Binomial Coefficient/Proof 1
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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
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:
- $\forall n \in \Z_{\ge 0}: \ds \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x} = \dfrac 1 {x \binom {n + x} n}$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $48$