Reciprocal as Summation of Binomial Coefficients of Reciprocals

Theorem

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

where $\dbinom {n - 1} k$ denotes a binomial coefficient.

That is, for example:

$\ds \dfrac 1 1$

$=$

$\ds 1$

$\ds \dfrac 1 2$

$=$

$\ds 1 - \dfrac 1 2$

$\ds \dfrac 1 3$

$=$

$\ds 1 - 2 \times \dfrac 1 2 + \dfrac 1 3$

$\ds \dfrac 1 4$

$=$

$\ds 1 - 3 \times \dfrac 1 2 + 3 \times \dfrac 1 3 - \dfrac 1 4$

$\ds \dfrac 1 5$

$=$

$\ds 1 - 4 \times \dfrac 1 2 + 6 \times \dfrac 1 3 - 4 \times \dfrac 1 4 + \dfrac 1 5$

$\ds \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \dbinom {n - 1} k \dfrac 1 {k + 1}$

$=$

$\ds \frac 1 n \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \binom n {k + 1}$

$\ds $

$=$

$\ds \frac 1 n \sum_{k \mathop = 1}^n \paren {-1}^{k + 1} \binom n k$

$\ds $

$=$

$\ds \frac 1 n \paren {\sum_{k \mathop = 0}^n \paren {-1}^{k + 1} \binom n k + 1}$

$\paren {-1}^1 \dbinom n 0 = -1$

$\ds $

$=$

$\ds \frac 1 n \paren {-\paren {1 - 1}^n + 1}$

$\ds $

$=$

$\ds \frac 1 n$

$\blacksquare$