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Assume $p$ is a prime number such that $p\equiv 1 \pmod3$, and $q=\lfloor \frac{2p}{3}\rfloor$.

If: $\frac{1}{1\cdot2} +\frac{1}{3\cdot4} +\cdots+\frac{1}{(q-1)\cdot q} =\frac{m}{n}$

For some integers $m,n$, what is the proof that $p\mid m$

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    @SimonMarkett $q-1$ is odd since $p$ is of the form $3k+1$, so $q = \lfloor (6k+2)/3 \rfloor = \lfloor 2k+2/3 \rfloor= 2k.$2012-08-25

1 Answers 1

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Let $p=3k+1$ and $H_n$ denote the n-th harmonic number. Since $\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ (1) the sum in the question is $\displaystyle \sum_{i=1}^q \frac{ (-1)^{i+1} }{i} = H_{2k}-H_k.$

Working in mod $p$, we add $p$ to the denominator of each term in $H_k$ : $ H_{2k} - H_k \equiv H_{2k} + \sum_{i=1}^k \frac{1}{p-i} = H_{p-1} =\frac{1}{2}\sum_{i=1}^{p-1} \left( \frac{1}{i} + \frac{1}{p-i}\right)\equiv \frac{1}{2}\sum_{i=1}^{p-1} \left( \frac{1}{i} + \frac{1}{-i}\right)=0. $

proving the result.


Using (1) we see that $\frac{1}{1\cdot2} +\frac{1}{3\cdot4} +\cdots+\frac{1}{(q-1)\cdot q} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} \cdots + \frac{1}{q-1} - \frac{1}{q}.$

The series is almost like a Harmonic number but it has alternating signs, so we add and subtract 2*even terms:

$\frac{1}{1} - \frac{1}{2} + \frac{1}{3} \cdots + \frac{1}{q-1} - \frac{1}{q} = \left( 1+ \frac{1}{2} + \cdots + \frac{1}{q} \right) -2 \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{q} \right).$

The factor of 2 cancels with the denominator of every term in the second bracket. Remembering that $q=2k$ gives $ \left( 1+ \frac{1}{2} + \cdots + \frac{1}{2k} \right) - \left( 1 + \frac{1}{2} + \cdots + \frac{1}{k} \right)= H_{2k} - H_k.$

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    @MohammedAl-mubark I've elaborated on getting to $H_{2k}-H_k.$ The easiest way to understand what I've written is to interpret everything as elements in the field $\mathbb{Z}_p$ and the fraction $\frac{1}{i}$ denotes the inverse of $i$ in that field. Do you know much about fields? If not, try googling "modulo arithmetic with fractions" and understanding that.2012-08-28