$1^{-1}+2^{-1}+\dots+\Big(\frac{p-1}{2}\Big)^{-1} \equiv -\frac{2^p - 2}{p} \mod p$ for an odd prime $p.$

Question

I've reduced a problem down to proving this identity. Unfortunately, I don't know where to even start. There has to be some way of expanding the RHS or combining terms on the LHS, but I don't see it. Any hints?

Answer 1

David W. Boyd writes:

There is a relationship between values of certain $H_n$ and Fermat's quotient $q_a = {(a^{p-1}-1)}/p \bmod p$. For example, a result from Eisenstein from 1850 [Dickson 1952, p. 41] states that $H_{(p-1)/2} \equiv -2q_2 \mod p$; this is easily seen from the binomial expansion of ${(1 + 1)}^p$.

This is your hint, and if that's not enough, the reference where to find a proof.

[Dickson 1952]: L. E. Dickson, History of the Theory of Numbers.
[Boyd 1994]: David W. Boyd, A $p$-adic Study of the Partial Sums of the Harmonic Series.

Answer 2

This is not true, for example $\text{p}=47$. We get:

$$\sum_{\text{k}=1}^{23}\frac{1}{\text{k}}=\frac{444316699}{118982864}\tag1$$

Which is not congruent to:

$$\frac{2-2^{47}}{47}\mod{47}=-2994414645858\mod{47}=6\tag2$$

Answer 3

This is an additional part to the given answer by @orlp, The following theorem discovred by many Authors repeatedly ( A fried -mann and J Tamarkine ,M Lerch and Nielson ], for $p \geq3$ and $1\leq m \leq p-2 $

$\sum_{j=1}^{p-1} j^m q_p(j) \equiv \frac{-B_m}{m} \bmod{p}$, here $q_p(j) =\frac{j^{p-1}-1}{p}$ denote the fermat quotient and $B_m$ are Bernouli numbers such that $p\nmid j$, For reference you may check this Handbook (page 548)