Wolstenholme’s theorem

We want to show first that the harmonic number

$$H_n=:\mathrm{\hspace{0.33em}1}+\frac{1}{2}+\frac{1}{3}+\mathrm{\dots }+\frac{1}{n}={\int }_0^1\frac{1-x^n}{1-x}dx$$

is never an integer ($n>1$).

Denote by $p$ the greatest prime number not exceeding $n$.  By Bertrand’s postulate there is a prime $q$ with .  Therefore we have  .  If $H_n$ were an integer, then the sum

$$n!H_n=\sum _{i=1}^n\frac{n!}{i}$$

had to be divisible by $p$.  However its addend ${\displaystyle \frac{n!}{p}}$ is not divisible by $p$ but all other addends are, whence the sum cannot be divisible by $p$.  The contradictory situation means that $H_n$ is not integer when  $n>1$.

Theorem (Wolstenholme).  If $p$ is a prime number greater than 3, then the numerator of the harmonic number

$$H_{p-1}=\mathrm{\hspace{0.33em}1}+\frac{1}{2}+\frac{1}{3}+\mathrm{\dots }+\frac{1}{p-1}$$

is always divisible by $p^2$.

Proof.  Consider the polynomial

$$f(x)=:(x-1)(x-2)\mathrm{\cdots }(x-p+1).$$

One has

$f(0)=(p-1)!=f(p)$(1)

and

$f(x)=x^{p-1}+a_1{x}^{p-2}+a_2{x}^{p-3}+\mathrm{\dots }+a_{p-2}x+(p-1)!$(2)

where $a_1,a_2,\mathrm{\dots },a_{p-2}$ are integers.  Because $1,\mathrm{\hspace{0.17em}2},\mathrm{\dots },p-1$ form a set of all modulo $p$ incongruent roots of the Fermat’s congruence (http://planetmath.org/FermatsTheorem)  $x^{p-1}\equiv 1(modp)$,  one may write the identical congruence

$x^{p-1}-1\equiv x^{p-1}+a_1{x}^{p-2}+a_2{x}^{p-3}+\mathrm{\dots }+a_{p-2}x+(p-1)!(modp).$(3)

It may be written by Wilson’s theorem  $(p-1)!\equiv -1(modp)$  as

$a_1{x}^{p-2}+a_2{x}^{p-3}+\mathrm{\dots }+a_{p-2}x\equiv \mathrm{\hspace{0.33em}0}(modp),$(4)

being thus true for any integer $x$.  From (4) one can successively infer that $p$ divides all coefficients $a_i$, i.e. that (4) actually is a formal congruence.

For the derivative of the polynomial $f(x)$ one has

$$f^{\prime }(x)=(x-2)\mathrm{\cdots }(x-p+1)+\mathrm{\dots }+(x-1)\mathrm{\cdots }(x-p+2)$$

and thus

$f^{\prime }(0)=-2\cdot 3\mathrm{\cdots }(p-1)-\mathrm{\dots }-1\cdot 2\mathrm{\cdots }(p-2).$(5)

The Taylor series (Taylor polynomial) of $f(x)$ coincides with $f(x)$:

$$f(x)=f(0)+\frac{f^{\prime }(0)}{1!}x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\mathrm{\dots }+\frac{f^{(p-1)}(0)}{(p-1)!}{x}^{p-1}$$

By (1), this equation implies

$0=f^{\prime }(0)+{\displaystyle \frac{f^{\prime \prime }(0)}{2}}p+\mathrm{\dots }+{\displaystyle \frac{f^{(p-1)}(0)}{(p-1)!}}{p}^{p-2}$(6)

Since  $p\mid a_{p-3}=\frac{f^{\prime \prime }(0)}{2}$,  one has  $p\mid f^{\prime \prime }(0)$.  It then follows by (6) that  $p^2\mid f^{\prime }(0)$.  And since (5) divided by $-(p-1)!$ gives

$$1+\frac{1}{2}+\frac{1}{3}+\mathrm{\dots }+\frac{1}{p-1}=-\frac{f^{\prime }(0)}{(p-1)!},$$

the assertion has been proved.