Reciprocal as Summation of Binomial Coefficients of Reciprocals
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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:
\(\ds \dfrac 1 1\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \dfrac 1 2\) | \(=\) | \(\ds 1 - \dfrac 1 2\) | ||||||||||||
\(\ds \dfrac 1 3\) | \(=\) | \(\ds 1 - 2 \times \dfrac 1 2 + \dfrac 1 3\) | ||||||||||||
\(\ds \dfrac 1 4\) | \(=\) | \(\ds 1 - 3 \times \dfrac 1 2 + 3 \times \dfrac 1 3 - \dfrac 1 4\) | ||||||||||||
\(\ds \dfrac 1 5\) | \(=\) | \(\ds 1 - 4 \times \dfrac 1 2 + 6 \times \dfrac 1 3 - 4 \times \dfrac 1 4 + \dfrac 1 5\) |
Proof
\(\ds \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \dbinom {n - 1} k \dfrac 1 {k + 1}\) | \(=\) | \(\ds \frac 1 n \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \binom n {k + 1}\) | Factors of Binomial Coefficient | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 n \sum_{k \mathop = 1}^n \paren {-1}^{k + 1} \binom n k\) | Translation of Index Variable of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 n \paren {\sum_{k \mathop = 0}^n \paren {-1}^{k + 1} \binom n k + 1}\) | $\paren {-1}^1 \dbinom n 0 = -1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 n \paren {-\paren {1 - 1}^n + 1}\) | Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 n\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $35$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $35$