# For some integers $m,n$,proving that $p\mid m$

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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I changed the dots to multiplications, I hope that is what you ment. – Simon Markett Aug 25 '12 at 11:08
Also is it clear that $q-1$ is odd, or could the sum possibly end with $\frac{1}{q(q+1)}$? – Simon Markett Aug 25 '12 at 11:11
@SimonMarkett $q-1$ is odd since $p$ is of the form $3k+1$, so $q = \lfloor (6k+2)/3 \rfloor = \lfloor 2k+2/3 \rfloor= 2k.$ – Ragib Zaman Aug 25 '12 at 11:14

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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Great answer! I'd prefer to finish $H_{p-1}=\sum_{i\in(\mathbf Z/p\mathbf Z)^\times}i^{-1}=\sum_{i\in(\mathbf Z/p\mathbf Z)^\times}i=0\in\mathbf Z/p\mathbf Z$, but that really just a matter of taste. – Marc van Leeuwen Aug 25 '12 at 15:16
@Marc That is a very nice proof of that fact, I wish I'd thought of that! – Ragib Zaman Aug 25 '12 at 15:30
Is that mean $m=0$ ? – Frank Aug 26 '12 at 5:45
@MohammedAl-mubark It means $m=0$ working mod $p$, which means $m$ is divisible by $p.$ – Ragib Zaman Aug 26 '12 at 6:41
Yes , i understand what you are saying,may be i should think before writing anything . – Frank Aug 26 '12 at 6:58