# Divisibility of Numerator of Sum of Sequence of Reciprocals

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## Theorem

Let $p$ be a prime number such that $p > 3$.

Consider the sum of the finite sequence of reciprocals as follows:

- $S = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 {p - 1}$

Let $S$ be expressed as a fraction in canonical form, that is:

- $S = \dfrac a b$

where $a$ and $b$ are coprime.

Then:

- $p^2 \divides a$

where $\divides$ denotes divisibility.

## Proof

### Lemma

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

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

where $\dbinom n k$ denotes a binomial coefficient.

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## Sources

- 1862: Joseph Wolstenholme:
*On certain properties of prime numbers*(*Quart. J. Pure Appl. Math.***Vol. 5**: pp. 35 – 39)